NCERT Solutions Class 8 Mathematics

NCERT syllabus introduces Class 8 students to several new topics and chapters. The students learn new Mathematical formulas and concepts at this level. And for this, they need the right learning aids and guidance to understand the chapters thoroughly. According to academic experts, NCERT Solutions are the best study materials to help the students. The solutions present detailed explanations of the Mathematics problems followed by examples. 

The subject-matter experts draft the Mathematics chapter-wise solutions in separate files to ease the study process for the students. Students can download the NCERT solutions for class 8 Mathematics from the official website of Extramarks for free.

 

Download NCERT Class 8 Mathematics Solutions 

Extramarks brings NCERT solutions to the class 8 students to impart a comprehensive knowledge about the subsequent chapters and topics. The study materials help the students in their examination preparation and during the daily revision process. Class 8 Mathematics NCERT solutions encourage the students to learn the subject further with more confidence and zeal. Students can take the help of the Mathematics solutions to develop their subject base and get high grades in the examination. The solutions cover answers to all the questions provided in the textbooks. The syllabus follows the NCERT curriculum, and the questions have been arranged as per the CBSE board guidelines. The students learn about the marks’ weightage, frequently asked questions, and significant chapters from the solutions. Advanced knowledge of the mark division helps the students to answer the questions to the point. The answers provided in the solutions teach the students how to answer the questions based on marks’ weightage. 

The NCERT solutions aid the students’ learning process and enable them to solve complex Mathematics problems with ease. Mathematics faculties all over the country recommend the 8th standard CBSE board students to study Mathematics with reference to NCERT solutions for class 8 Mathematics to develop a solid grasp of the subject. The students can avail these accurate and reliable Mathematics solutions for free from Extramarks. For this, they need to sign in to the website and access the materials. The class 8 students can enjoy the convenience to download the solutions online and study them later.

 

NCERT Solutions for Class 8 Mathematics Chapter Wise Download

NCERT Mathematics book class 8 solutions are available on Extramarks website. Students can easily find the chapter-wise solutions they need to study. The systematic arrangement of the chapters in the solution book saves time. The solutions provided are based on the NCERT books syllabus. The explanations accompanied by examples guide the students to understand the core concept of each chapter. Students find even the difficult and complex topics elaborated and decoded in a simpler manner. The use of easy-to-understand language enables the students to learn the crux of every chapter. The direct and logical approach of the solutions encourages the students to solve more Mathematical problems and practice consistently. Understanding one chapter eases the learning process of others as almost all the chapters are interlinked with each other. 

 

NCERT Solutions for Class 8 Mathematics – Chapter-wise Solutions

 

Chapter 1: Rational Numbers

1.1 – Introduction 

1.2 – Properties of Rational Numbers 

1.3 – Representation of Rational Numbers on the Number Line

1.4 – Rational Number between Two Rational Numbers 

 

Chapter 2: Linear Equations in One Variable 

2.1 – Introduction

2.2 – Solving Equations Which have Linear Expressions on One Side and Numbers on the Other Side

2.3 – Some Applications 

2.4 – Solving Equations Having the Variables on Both Sides

2.5 – Some More Application Names

2.6 – Reducing Equations to Simpler Form

2.7 – Equations Reducible to the Linear Form

 

Chapter 3: Understanding Quadrilaterals 

3.1 – Introduction 

3.2 – Polygons

3.3 – Some of the Measures of the Exterior Angles of a Polygon

3.4 – Kinds of Quadrilaterals 

3.5 – Some Special Parallelograms 

 

Chapter 4: Practical Geometry

4.1 – Introduction 

4.2- Constructing a Quadrilateral

4.3 – Some Special Cases

 

Chapter 5: Data Handling

5.1 – Looking for Information 

5.2 – Organizing Data

5.3 – Grouping Data

5.4 – Circle Graph or Pie Chart

5.5 – Chance and Probability 

 

Chapter 6: Squares and Square Roots

6.1 – Introduction

6.2 – Properties of Square Numbers 

6.3 – Some More Interesting Numbers 

6.4 – Finding the Square of a Number

6.5 – Square Roots

6.6 – Square Roots of Decimals

6.7 – Estimating Square Root

 

Chapter 7: Cubes and Cube Roots

7.1 – Introduction

7.2 – Cubes 

7.3 – Cubes Roots

 

Chapter 8: Comparing Quantities

8.1 – Recalling Ratios and Percentages 

8.2 – Finding the Increase and Decrease Percentage

8.3 – Finding Discounts 

8.4 – Prices Related to Buying and Selling (Profit and Loss)

8.5 – Sales Tax/Value Added Tax/Goods and Services Tax

8.6 – Compound Interest

8.7 – Deducing a Formula for Compound Interest

8.8 – Rate Compounded Annually or Half Yearly (Semi-Annually)

8.9 – Applications of Compound Interest Formula

 

Chapter 9: Algebraic Expressions and Identities 

9.1 – What are Expressions?

9.2 – Terms, Factors, and Coefficients 

9.3 – Monomials, Binomials, and Polynomials 

9.4 – Like and Unlike Terms

9.5 – Addition and Subtraction of Algebraic Expressions 

9.6 – Multiplication of Algebraic Expressions: Colon

9.7 – Multiplying a Monomial by a Monomial 

9.8 – Multiplying a Monomial by a Polynomial

9.9 – Multiplying a Polynomial by a Polynomial

9.10 – What is an Identity

9.11 – Standard Identities 

9.12 – Applying Identities 

 

Chapter 10: Visualising Solid Shapes

10.1 – Introduction

10.2 – View of 3D-Shapes

10.3 – Mapping Space Around Us

10.4 – Faces, Edges, and Vertices 

 

Chapter 11: Mensuration 

11.1 – Introduction

11.2 – Let Us Recall

11.3 – Area of Trapezium

11.4 – Area of General Quadrilateral

11.5 – Area of Polygons

11.6 – Solid Shapes

11.7 – Surface Area of Cube, Cuboid, and Cylinder

11.8 – The volume of Cube, Cuboid, and Cylinder

11.9 – Volume and Capacity 

 

Chapter 12: Exponents and Powers

12.1 – Introduction 

12.2 – Powers with Negative Exponents

12.3 – Laws of Exponents

12.4 – Use of Exponents to Express Small Numbers in Standard Forms

 

Chapter 13: Direct and Inverse Proportions 

13.1 – Introduction

13.2 – Direct Proportion 

13.3 – Inverse Proportion 

 

Chapter 14: Factorization 

14.1 – Introduction 

14.2 – What is Factorization?

14.3 – Division of Algebraic Expressions 

14.4 – Division of Algebraic Expressions Continued (Polynomial/Polynomial)

14.5 – Can You Find the Error?

 

Chapter 15: Introduction to Graphs 

15.1 – Introduction 

15.2 – Linear Graphs

15.3 – Some Applications 

 

Chapter 16: Playing with Numbers 

16.1 – Introduction 

16.2 – Numbers in General Form

16.3 – Game with Numbers 

16.4 – Letters for Digits 

16.5 – Test of Divisibility 

 

NCERT Class 8  Chapter-wise Solutions in Hindi

 

अध्याय 1 – परिमेय संख्याएं

1.1 – परिमेय संख्याओं के गुणों पर प्रश्न।

1.2 – एक संख्या रेखा पर परिमेय संख्याओं को निरूपित करने वाले प्रश्न।

अध्याय 2 – एक चर में रैखिक समीकरण

2.1 – एक चर में रैखिक समीकरणों को सरल और हल करने पर प्रश्न।

2.2 – रैखिक समीकरणों के अनुप्रयोग पर प्रश्न।

2.3 – कुछ जटिल रैखिक समीकरणों को सरल बनाने से संबंधित प्रश्न।

2.4 – दैनिक जीवन में एक चर में रैखिक समीकरणों के अनुप्रयोग पर प्रश्न।

2.5 – एक समीकरण को रैखिक में कैसे परिवर्तित करें और उन्हें कैसे हल करें, इस पर प्रश्न।

2.6 – विविध प्रश्न।   

 

अध्याय 3 – चतुर्भुज को समझना

3.1 – एक बंद आकृति के आंतरिक और बाहरी भागों, बहुभुज के विकर्ण, बहुभुज के कोण योग गुण, और अवतल और उत्तल बहुभुज पर आधारित प्रश्न।

3.2 – बहुभुज के बहिष्कोणों के मापों के गुणधर्म पर प्रश्न।

3.3 – समांतर चतुर्भुज, पतंग, समलम्ब चतुर्भुज जैसे विभिन्न प्रकार के चतुर्भुजों पर प्रश्न।

3.4 – आयत, वर्ग और समचतुर्भुज जैसे विशिष्ट प्रकार के चतुर्भुजों पर प्रश्न। 

अध्याय 4 – व्यावहारिक ज्यामिति

4.1 – एक चतुर्भुज की रचना पर प्रश्न जब चार भुजाएँ और एक विकर्ण दिया गया हो।

4.2 – त्रिभुज की रचना पर प्रश्न जब तीन भुजाएँ और एक विकर्ण दिया गया हो।

4.3 – एक चतुर्भुज की रचना पर प्रश्न जब तीन कोण और दो आसन्न भुजाएँ दी गई हों।

4.4 – एक चतुर्भुज की रचना पर प्रश्न जब तीन भुजाएँ और दो कोण दिए गए हों।

4.5 – अन्य गुण ज्ञात होने पर एक चतुर्भुज की रचना पर प्रश्न। 

अध्याय 5 – डेटा हैंडलिंग

5.1 – हिस्टोग्राम पर प्रश्न।

5.2 – पाई चार्ट या वृत्त ग्राफ पर प्रश्न।

5.3 – परिणामों, घटनाओं, घटनाओं की कुल संख्या और कुल अनुकूल घटनाओं सहित संभाव्यता पर प्रश्न। 

अध्याय 6 – वर्ग और वर्गमूल

6.1 – क्रमागत योग विधि की हिट और परीक्षण विधि का उपयोग करके किसी संख्या के वर्ग की गणना करने पर प्रश्न।

6.2 – द्विपद प्रसार सूत्र द्वारा किसी संख्या के वर्ग की गणना करने या संख्या को स्वयं से गुणा करने और पाइथागोरस त्रिक से संबंधित प्रश्न।

6.3 – क्रमागत घटाव या अभाज्य गुणनखंड विधि द्वारा किसी संख्या के वर्गमूल की गणना पर प्रश्न।

6.4 – एक दशमलव संख्या के वर्गमूल की गणना के साथ-साथ अनुमान या भाग विधि द्वारा वर्गमूल की गणना पर प्रश्न।

अध्याय 7 – घन और घनमूल

7.1 – अभाज्य गुणनखंडों द्वारा किसी संख्या का घन ज्ञात करने या क्रमागत विषम संख्याओं के योग से संबंधित प्रश्न।

7.2 – अभाज्य गुणनखंड विधि द्वारा घनमूलों पर आधारित प्रश्न।

अध्याय 8 – मात्राओं की तुलना

8.1 – अनुपात और प्रतिशत पर आधारित प्रश्न और उनके व्यावहारिक अनुप्रयोग।

8.2 – प्रतिशत में वृद्धि और कमी, लाभ और हानि, छूट, वैट और बिक्री कर में प्रश्न।

8.3 – साधारण और चक्रवृद्धि ब्याज पर प्रश्न।

अध्याय 9 – बीजीय व्यंजक और सर्वसमिकाएँ

9.1 – बीजीय व्यंजकों और समान और विषम पदों को जोड़ने और घटाने पर आधारित प्रश्न।

9.2 – बीजीय व्यंजकों के गुणन पर आधारित प्रश्न।

9.3 – एक एकपदी से एक बहुपद के गुणन से संबंधित प्रश्न।

9.4 – एक बहुपद को दूसरे बहुपद से गुणा करने वाले प्रश्न।

9.5 – सर्वसमिकाओं पर आधारित प्रश्न।

अध्याय 10 – ठोस आकृतियों की कल्पना करना

10.1 – एक 3-डी वस्तु को विभिन्न दृष्टिकोणों से आलेखित करने पर प्रश्न।

10.2 – मानचित्रण रिक्त स्थान पर प्रश्न।

10.3 – किसी 3D वस्तु के कोनों, फलकों, शीर्षों और किनारों को ढूँढना।

अध्याय 11 – क्षेत्रमिति

11.1 – बुनियादी क्षेत्रमिति प्रश्न।

11.2 – चतुर्भुज जैसे चतुर्भुज के क्षेत्रों पर आधारित प्रश्न।

11.3 – एक ठोस आकृति के पृष्ठीय क्षेत्रफल की गणना करना।

11.4 – आकृतियों के आयतन की गणना पर प्रश्न।

अध्याय 12 – घातांक और शक्तियाँ

12.1 – घातांक के नियमों पर आधारित प्रश्न।

12.2 – मानक रूप या वैज्ञानिक रूप पर आधारित प्रश्न और घातांक का उपयोग करके किसी संख्या को उनमें कैसे परिवर्तित किया जाए।

अध्याय 13 – प्रत्यक्ष और व्युत्क्रमानुपाती

13.1 – प्रत्यक्ष अनुपात पर आधारित प्रश्न।

13.2 – प्रतिलोम अनुपात पर आधारित प्रश्न और ऐसे प्रश्न जिनमें प्रत्यक्ष और अप्रत्यक्ष अनुपात शामिल हों।

अध्याय 14 – गुणनखंड

14.1 – कारक बनाने के लिए पुनर्समूहीकरण शर्तों पर आधारित प्रश्न।

14.2 – एक बीजीय व्यंजक को बहुपद या एकपदी से विभाजित करने में आने वाली समस्याएँ।

14.3 – एक बहुपद को दूसरे बहुपद से भाग देने में आने वाली समस्याएँ।

14.4 – प्रश्न समीकरणों में त्रुटियों को कवर कर रहे हैं और समीकरणों और पहचानों को सही साबित करने के लिए उन्हें हटा रहे हैं।

अध्याय 15 – रेखांकन का परिचय

15.1 – रेखा आलेखों पर प्रश्न।

15.2 – x और y-अक्ष से संबंधित समस्याएं, और समन्वय प्रणाली और उन्हें एक ग्राफ पेपर पर प्लॉट करना।

15.3 – दो चरों का उपयोग करते हुए आलेख और दो चरों में रैखिक समीकरणों के अनुप्रयोग पर प्रश्न।

 

अध्याय 16 – संख्याओं के साथ खेलना

16.1 – जोड़, घटाव, गुणा और भाग पर प्रश्न और पहेलियाँ।

16.2 – किसी संख्या की 2, 3, 5, 9 और 10 से विभाज्यता का परीक्षण करना।

 

NCERT Class 8 Mathematics Solutions Chapter 1: Rational Numbers Exercise 

NCERT Class 8 Mathematics Solutions on Chapter 1: Students learn about natural and whole numbers and the properties of integers. NCERT solutions for class 8 Mathematics explain the representation of rational numbers on the number line and the role of zero and one in multiplication and addition. 

The first topic of the chapter is an introduction to rational numbers. The introduction is one constitutive section to obtain the fundamentals of the respective chapter. The first exercise deals with properties and how to use them the right way. The students learn how to represent numbers on the number line in the next topic. The third and last topic of the chapter teaches how to find a rational number between two rational numbers. 

 

NCERT Class 8 Mathematics Solutions Chapter 2: Linear Numbers in One Variable Exercise

Chapter 2 introduces the class 8 students the concept of linear equations in one variable. The solution guides the students with several methods and examples to ease the Mathematics solving process. The exercises at the end of the chapter help students assess their understanding and ability to solve more complex sums. 

In the beginning, the topic introduces the students to the linear equation and its forms. The students learn to solve equations with one and two variables. The second topic deals with methods of solving equations. The third topic explains the application of linear equations. The students learn about the application methods. The examples preferred in the solutions benefit the students. The next concept teaches to reduce equations to simpler forms, how to reduce equations, and solve the exercise questions. The students also learn how to reduce the equations to linear forms. NCERT solutions are the best material to practise the additional questions. The students can boost their overall score by securing good marks in this section. 

 

NCERT Class 8 Mathematics Solutions Chapter 3: Understanding Quadrilaterals Exercise 

NCERT Class 8 Mathematics Solution chapter 3 provide the students with a comprehensive understanding of quadrilaterals and the study of their properties. Students learn about different polygons such as hexagons, pentagons, octagons,  and so on.

The students learn about the formulas of interior angles, areas, and perimeter. The students are suggested to solve the in text and end text questions in the exercises , to develop a profound grasp of the concept.

The next topic includes the exterior angles of a polygon. The students learn about exterior angles and the sums and solve the questions. The examples help them to understand the concept clearly. Class 8 students also learn about the properties of various shapes like  parallelogram, trapezium, and also that of a kite. 

The last topic describes the special parallelograms, including squares, rectangles, and rhombus. Students learn about two and three-dimensional (3D) shapes. Practising the additional questions of every topic helps the students clear all doubts regarding each subtopic.

 

NCERT Class 8 Mathematics Solutions Chapter 4: Practical Geometry Exercise 

As the class 8 students have learned about quadrilaterals in detail, chapter 4 teaches the students: how to form quadrilaterals based on the given parameters like lengths of the sides, angles, length of diagonals, etc. The students can excel in solving complex problems with regular practice of the topics. 

The chapter ‘Practical Geometry’ comprises three exercises. The students learn about constructing quadrilaterals. Solving examples and additional questions from NCERT Solution books help the students acquire a good grasp of the chapter. The students can practice and revise the steps regularly to strengthen their ability to construct required shapes with the exact measurements given.

The last topic teaches the students how to construct different special shapes? The class 8 students learn how to draw a rhombus. Geometry is a scoring topic, and students can improve their overall results with regular practice. 

 

NCERT Class 8 Mathematics Solutions Chapter 5: Data Handling Exercise

In this Chapter, the students learn how to organize the data systematically. Class 8 students learn how to represent information or data through bar graphs, pie charts, or histograms. The chapter carries vital importance in the Mathematics syllabus of CBSE Class 8 students. 

The chapter includes five chapters, and the students need to learn all. The chapter explains how to handle and use data. They learn the methods and check the elaborate examples to understand the topic to the core. . The chapter includes concepts like why to organise group data and their advantages and disadvantages. The students must solve the exercise questions and sample papers to learn about lower-class limits, upper-class limits, and how to make a pie chart?

The students learn the uses of pie charts with the advantages and disadvantages. They can improve their score by practicing the exercise questions and examples. The last topic describes chance and probability. Students learn about probability formulas and chance and probability of events, besides learning how to represent through various graphs. Examples in NCERT Mathematics book class 8 solutions enable the students to create a strong base in data handling. 

 

NCERT Class 8 Mathematics Solutions Chapter 6: Squares and Square Roots Exercise

‘Square and Square Roots’ is one of the most crucial chapters in the class 8 Mathematics syllabus. The chapter teaches the students the concept of square and square roots and the pattern which resembles the same. When the students become familiar with the concepts they find it more exciting to solve the exercises. The chapter includes the concept and properties of Pythagorean triples. 

The chapter covers seven exercises. The students must practise them all to attain a solid grasp of the concept. It is essential to learn about square numbers and the square root properties. 

The second chapter is about how to find the square of a number. The students learn about even and odd square numbers, their properties, and square roots. NCERT Mathematics Class 8 solutions guide the students to speed up their preparation for the examination.

The next topic in Chapter 6 is how to find the square root of a number by repetitive subtraction. The students learn how to solve the Mathematics problems and practice more examples and questions associated with the topic. The chapter also teaches the students how to find the square root of a decimal. 

The last topic is about estimating a square root. The references and solutions help the students learn how to measure a square root. The topic is an important one from the examination perspective. 

The students can easily get a high score in this chapter with guided practice.  All long and short questions come from this chapter. The class 8 students need to practice the solutions to delve deep into the concept.

 

NCERT Class 8 Mathematics Solutions Chapter 7: Cubes and Cubes Roots Exercise

Chapter 7 in the CBSE Class 8 Mathematics syllabus is one continuation of the previous topic of ‘Square and Square Roots’. It introduces the students to the concepts of cubes and cube roots for the first time. Class 8 students learn about patterns and properties associated with the same as they proceed with the chapter. The students are suggested to solve and practise the exercises to get complete knowledge of the chapter.

The first topic of the chapter covers the definition of a cube and how to find perfect cube numbers. The students learn about the application of perfect cubes and practice questions associated with the concept. It is necessary to solve the questions to understand the topic in detail. 

Another significant topic is the cube root of a number. In this subtopic, the students learn about cube roots and how to find the cube root of a number. The NCERT solutions by the subject experts are of great benefit to the students to understand the topic in-depth.

Mathematics faculties recommend the students of class 8 study the cube and cube roots table. The chapter is not a big one. Six marks are allotted for the chapter in the examination, but it does not mean that the students can skip this chapter. 

Another topic includes the cube root by prime factorization method. The students learn how to find the cube root of a number. Only the formulas change. A few examples include the cube roots of 759, 512, 216, and so on. The students must practice the samples to acquire detailed knowledge about the chapter.

 

NCERT Class 8 Mathematics Solutions Chapter 8: Comparing Quantities Exercise

The chapter ‘Comparing Quantities’ is an absorbing topic for the class 8 students because they learn about percentages (%), sales, discounts, and other real-life commercial terms. We come across these words often in our everyday life. Students learn to calculate simple and compound interests, which are of daily use in our lives. The students need to practise the solutions and exercises to gain a solid grasp of the topics. Knowledge regarding these topics helps the students. to get an experiential learning of the concept.

The students find the concept to be easy to learn. It comprises nine exercises, and each one is equally significant to understanding the chapter in its entirety. The chapter holds 16 marks in the examination, and therefore the students must prepare the topics comprehensively for the exams. 

The first topic is percentages and ratios. The students learn about differences in percentages and increase and decrease in percentage. It is essential to memorise the formulas to solve the questions at ease. The topic includes how to convert the fraction into percentages, the difference between percentages and percent, etc. NCERT Mathematics book class 8 solutions explain all the topics to the students with a direct approach. 

Class 8 students study and learn the formulas on profit and loss, their percentages, discounted percentages, with rise and fall in it. The students need to practise the percentage formulas and solve the questions; using the formulae.

Another topic explains simple interest and compound interest. The students get to know about the formulas associated with the topics. Then, another topic covers the rate compounded half-yearly and annually. It is a crucial one. Practice is the only way to have a grasp over it. 

The last topic is about applications of compound interest formulas. The students learn about the advantages and disadvantages of compound interest. The students should solve questions and exercise as many as they can to be at ease with the concept and solve the Mathematics problems effortlessly. 

 

NCERT Class 8 Mathematics Solutions Chapter 9: Algebraic Expressions and Identities Exercise 

Class 8 is one of the initial academic years when the students are introduced to the basics of Algebra. Students become familiar with the terms related to Algebraic Expressions and Identities. The chapter explains the topics such as factors, monomials, binomials, polynomials, coefficients, and so on. Students learn how to solve Mathematics  problems based on these formulas. The students can acquire a solid grasp of the chapter through  regular practice of the exercises and solutions.

The chapter comprises twelve exercises. The first topic of the chapter is Algebraic expressions. In this chapter, the students learn about different kinds of Algebraic expressions and their formulas. The students must solve the NCERT solutions to understand the topic extensively. Some of the most significant expressions include monomial, binomial, polynomial, and variable expressions. CBSE Class 8 students have to read all the expressions and the examples before solving the exercise questions.

The next topic is about factors, coefficients, and terms. The formulas and solved examples play a crucial role in the students’ learning process. The chapter includes the addition and subtraction of algebraic expressions. The 6th topic is about the multiplication of algebraic expressions. The students learn how to multiply algebraic expressions, the formulas, and the terms used in the chapter.

The chapter continues to teach the students how to multiply monomial with a monomial and monomial by a polynomial. Exercise 11 explains the standard identities, which are about the conditions for the standard equation. The students learn about every kind of standard identity in this chapter. 

The last topic describes how to apply identities. The students learn the formulas and methods to solve the exercises and practice the questions.  The chapter needs some more attention, as it is a lengthy one.

 

NCERT Class 8 Mathematics Solutions Chapter 10: Visualizing Solid Shapes Exercise 

Chapter 10 intrigues the class 8 students with its concept. The students are introduced to 3D Geometry. They learn how to visualize solid shapes in different dimensions. The chapter includes the topics like edges, faces, and vertices and teaches the students to create a solid 3D shape. The students learn Euler’s formula. The chapter needs the undivided attention of the students. The students can gather a comprehensive understanding of the topic, by solving the exercise questions.

The topic comprises four exercises. The concept is a new one, but the concept is easy to learn. The first topic is about mapping the space around us. The students learn how to space maps and their uses. Another crucial topic is about faces, edges, and vertices of 3-Dimensional shapes. The students also get to learn about types of polyhedrons. The chapter is not too long and easy to score from the examination point of view.

 

NCERT Class 8 Mathematics Solutions Chapter 11: Mensuration Exercise 

The chapter intends to develop the students’ ideas of quadrilaterals and triangles. It is about measuring the area of these figures. The students also get to learn to calculate the area and volumes of cubes and cuboidal structures with the use of various formulas. The students need to solve the in-text and end text questions to have a clear understanding of the topic.

The chapter has nine exercises, and 15 marks in the examination. The students learn about the area of a trapezium and its properties.

Class 8 students learn the formulas of the area and perimeter of the trapezium and practise the questions associated with it. Another vital topic is the derivation of the area of a trapezium and its applications of a trapezium. The students must practise the derivation exercises in detail as the topic is most likely to come in long answer questions. NCERT Mathematics book class 8 solutions help the students by offering them explanations and elaborate examples.

The next topic covers the area of a quadrilateral and the properties of quadrilaterals. The students need to learn and memorise the formulas to solve the questions at ease. Topic 5 is about the area of a polygon which teaches about the area of a polygon and its formula. 

Another topic includes the surface area of a cube, cuboid, and cylinder and how to find them. The students learn how to find the length of the edge of a cube. Regular practice is the most effective tool to remember formulas and apply them rightly while solving problems. . The chapter presents the formulas of the cube, cuboid, and cylinder to the students. NCERT Mathematics book class 8 solutions are of great help to the students to understand the topics in-depth.

The last topic of Chapter 11 talks about volume and capacity. The students learn the definition of volume,  capacity,  their differences, and formulas. The subtopics need profound attention from the students to understand the answers. 

 

NCERT Class 8 Mathematics Solutions Chapter 12: Exponents and Powers Exercise 

The chapter introduces the students to the concepts of power and exponents. Representation of numbers in standard form using exponents and comparisons is one intricate section of the chapter. The students need to solve the exercises to understand the concept with clarity. The chapter consists of four exercises. It teaches the students about exponents VS power and integers with positive and negative exponents. The students solve the examples associated with every topic and practise the questions.

The class 8 students also get to learn the integer exponent rules. The chapter is easy to learn. The students can understand the topics easily with little attention and dedication. Another topic is the laws of exponents. The students learn about the rules of the exponents and practise the examples to be familiar with the topic. It is essential to learn about the product and quotient with a similar base and solve the questions associated with each. Two other significant topics are the product to power and quotient to power.

The last and the final topic of the chapter is negative and fractional exponent rules. The students are advised to solve the exercise questions and practise the examples. The students learn about zero power. Another topic in the chapter discusses the use of exponents to express the small number in standard form. The students need to remember all the formulas and methods to solve the questions at ease.

 

NCERT Class 8 Mathematics Solutions Chapter 13: Direct and Inverse Proportions Exercise

Chapter 13 explains that two involved quantities can decrease or increase in a way that the corresponding values remain the same, and the opposite happens in inverse proportion. The students need to solve the exercise questions regularly for vivid memorization.

The chapter covers two exercises. The first exercise teaches about direct proportions. The students know what direct proportions are and the symbols, formulas and, constants of proportionality. The second topic is inverse variation. It teaches the students about the inverse variation, its equations and formulas. The students must solve the exercise questions to acquire a good grasp of the concept. NCERT solutions help the students with detailed explanations of the topics and exercise questions. 

 

NCERT Class 8 Mathematics Solutions Chapter 14: Factorization Exercise 

The chapter deals with the factorization of natural numbers and algebraic expressions. Class 8 students learn how to factorize using factorization identities, regrouping terms, etc. The chapter also teaches how to divide a polynomial or a monomial by a monomial. The students must practise the exercise questions from the factorization.

Chapter 14 carries a significant role in the class 8 Mathematics syllabus. It carries 12 marks. The students need to read and study the chapter thoroughly to understand the basics. The chapter consists of five exercises. The students learn how to factorize using common factors and practise the questions associated with the topic. 

The next topic deals with polynomial division and the algorithm of long division. It is a crucial one. For this reason, the students need to practise the examples and study the solved examples minutely. The students learn about the types of a polynomial division and the division of a polynomial by a polynomial. The last topic teaches how to find the error. The students can take the assistance of Class 8 Mathematics NCERT Solutions to understand the concepts better.

 

NCERT Class 8 Mathematics Solutions Chapter 15: Introduction to Graphs Exercise 

Class 8 syllabus has already introduced the concept of data handling to the students. In ‘Introduction to Graphs’, the students learn how to represent data in different types of graphs like histograms, pie charts, linear graphs, etc. The chapter teaches the students: how to locate points and sync them on the graph? 

The students can practise exercise questions and exercise to have a clear idea about the concept. Chapter 15 comprises three exercises. The chapter teaches the students about line graphs, types of line graphs, straight-line graphs, vertical line graphs, and double line graphs. The students learn the concepts through the intext and end text exercise for a better understanding. 

Another topic is line graphs, where the students learn about linear graph equations and solve the exercise questions. NCERT Mathematics book class 8 solutions enable the students to understand and absorb the topic better.

The last topic is the application of linear graphs. The students learn about the applications of linear graphs and solve the questions associated with it.

 

NCERT Class 8 Mathematics Solutions Chapter 16: Playing with Numbers Exercise 

The last chapter is the gist of all the topics allotted in the CBSE Class 8 Mathematics syllabus. The most crucial concepts have been discussed in the chapter in minute detail. It includes five exercises. The students need to use the methods and formulas to justify the tests of divisibility. 

The first topic teaches the students about numbers in a general form. The students learn how to represent numbers in general form and solve the exercise questions.

The second topic is numbers. The students learn the rules to solve puzzles and the exercise and understand the topic better. The last topic is the test of divisibility. The students learn all the divisibility rules. Solving questions is the best way to have a good grasp of the concept. The students can bank on the NCERT Solutions for thorough and complete preparation. 

 

CBSE Class 8 Mathematics Unit-Wise Marks Weightage 

Unit No Chapter Name Marks Weightage
Unit 1 Rational Number 8
Unit 2 Linear Equations in One Variable 10
Unit 3 Understanding Quadrilaterals 13
Unit 4 Practical Geometry 08
Unit 5 Data Handling 11
Unit 6 Squares and Square Roots 12
Unit 7 Cubes and Cube Roots 06
Unit 8 Comparing Quantities 16
Unit 9 Algebraic Expressions and Identities 12
Unit 10 Visualizing Solid Shapes 07
Unit 11 Mensuration 15
Unit 12 Exponents and Powers 06
Unit 13 Direct and Inverse Proportions 08
           

Unit 14

     

Factorisation

12
Unit 15 Introduction to Graphs 12
Unit 16 Playing with Numbers 10

 

Benefits of NCERT Solutions for Class 8 Mathematics by Extramarks

NCERT solutions are the most effective study materials for class 8 students. Subject-matter experts have compiled and drafted the materials to impart detailed knowledge about Mathematics to the students. They are provided with step-wise solutions for each chapter to make the students familiar with all the concepts  given  in the syllabus. The students are encouraged to solve more Mathematical questions with great confidence and ease. The solutions provided in the NCERT books are reliable and accurate.  The students are guided the right way, and are able to score good marks in the examination. 

The Class 8 students can enjoy several benefits of the NCERT Mathematics book class 8 Solutions in their learning process. The materials are accessible through any smart device that can connect to the internet. Extramarks believes in an open education system where the students can download the Solutions online and study them later, at their convenience. The answers are available on the website for free. 

In a nutshell  , the benefits include,

  • Expert Mathematics faculty with years of experience and knowledge have created the Solutions.
  • Solutions are both descriptive yet concise. .
  • The NCERT solutions strictly follow the CBSE guidelines.
  • The Solutions help the students practise the exercises and solve the questions.
  • It guides the students to make a solid preparation for the examination.
  • The solutions help the students learn about the mark weightage of every chapter and how to write on-point answers for each question.
  • The solutions allow the students to study the subject within the time frame.
  • The students can complete the syllabus on time with NCERT Solutions.

 

What are the Key Features of Extramarks Class 8 Mathematics NCERT Solutions?

The key features of Extramarks’ Class 8 Mathematics NCERT Solutions are:

  • Subject-matter experts have prepared the solutions.
  • The solutions are well-structured and self-explanatory.
  • Students can study  at their convenience, download specific chapter wise  topics  and prepare the important questions for the examination.
  • Students can revise  regularly before the exam.
  • The solutions are available on the site for free.
  • It helps the students get  a clear idea about the topics and concepts.
  • The students can  check improvement in their performance with constant revision and practice.
  • They can  devote more time on  difficult  and complex topics.

Conclusion:  One Solution for all problems

For the class 8 students who appear for the Mathematics examination, the NCERT solutions are very helpful. The solutions help the students to attain a solid grasp of Mathematics and perform well in the examination; irrespective of whether it is a class test or a final exam. NCERT Mathematics solutions are the best resources the students can avail. 

Q.1 Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube.

(i) 243 (ii) 256 (iii) 72 (iv) 675 (v) 100

Ans

(i)

The prime factorisation of 243 is as follows:324338132739331243=3×3×3¯×3×3

Here, two 3s are left which does not appear in a group of three. To make 243 a cube, one more 3 is required.
In that case, we have to multiply 243 by 3
i.e. 243 × 3 = 3 × 3 × 3 × 3 × 3 × 3 = 729, which is a perfect cube.
Hence, the smallest natural number by which 243 should be multiplied to make it a perfect cube is 3.

(ii)

The prime factorisation of 256 is as follows:225621282642322162824221256=2×2×2¯×2×2×2¯×2×2

Here, two 2s are left which does not appear in a group of three. To make 256 a cube, one more 2 is required.

If we multiply 256 by 2, we get

256 × 2 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 512, which is a perfect cube.

Hence, the smallest natural number by which 256 should be multiplied to make it a perfect cube is 2.

(iii)

The prime factorisation of 72 is as follows:2722362183933172=2×2×2¯×3×3

Here, two 3s are left which are not in a group of three. To make 72 a perfect cube, one more 3 is required.

Then, we obtain

72 × 3 = 2 × 2 × 2 × 3 × 3 × 3 = 216, which is a perfect cube.

Hence, the smallest natural number by which 72 should be multiplied to make it a perfect cube is 3.

(iv)

The prime factorisation of 675 is as follows:36753225375525551

675=3×3×3¯×5×5

Here, two 5s are left which are not in a group of three. To make 675 a cube, one more 5 is required.

If we multiply 675 by 5, we get

675 × 5 = 3 × 3 × 3 × 5 × 5 × 5 = 3375, which is a perfect cube.

Hence, the smallest natural number by which 675 should be multiplied to make it a perfect cube is 5.

(v)

The prime factorisation of 100 is as follows:2100250525551100=2×2×5×5

Here, two 2s and two 5s are left which are not in a triplet. To make 100 a cube, we require one more 2 and one more 5.

Then, we obtain

100 × 2 × 5 = 2 × 2 × 2 × 5 × 5 × 5 = 1000, which is a perfect cube

Hence, the smallest natural number by which 100 should be multiplied to make it a perfect cube is 2 × 5 = 10.

Q.2 Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube.

(i) 81 (ii) 128 (iii) 135 (iv) 192 (v) 704

Ans

(i)

The prime factorisation of 81 is as follows:38132739331

81 = 3 × 3 × 3 × 3

Here, one 3 is left which is not in a triplet.

If we divide 81 by 3, then it will become a perfect cube.

Therefore, 81 ÷ 3 = 27 = 3 × 3 × 3 is a perfect cube.

Hence, the smallest number by which 81 should be divided to make it a perfect cube is 3.

(ii)

The prime factorisation of 128 is as follows:21282642322162824221

128 = 2 × 2 × 2 × 2 × 2 × 2 × 2

Here, one 2 is left which is not in a group of three.

If we divide 128 by 2, then it will become a perfect cube.

Therefore, 128 ÷ 2 = 64 = 2 × 2 × 2 × 2 × 2 × 2 is a perfect cube.

Hence, the smallest number by which 128 should be divided to make it a perfect cube is 2.

(iii)

The prime factorisation of 135 is as follows:

3135345315551

135 = 3 × 3 × 3 × 5

Here, one 5 is left which is not in a group of three.

If we divide 135 by 5, then it will become a perfect cube.

Thus, 135 ÷ 5 = 27 = 3 × 3 × 3 is a perfect cube.

Hence, the smallest number by which 135 should be divided to make it a perfect cube is 5.

(iv)

The prime factorisation of 192 is as follows:

219229624822421226331

192 = 2 × 2 × 2 × 2 × 2 × 2 × 3

Here, one 3 is left which is not in a group of three.

If we divide 192 by 3, then it will become a perfect cube.

Thus, 192 ÷ 3 = 64 = 2 × 2 × 2 × 2 × 2 × 2 is a perfect cube.

Hence, the smallest number by which 192 should be divided to make it a perfect cube is 3.

(v)

The prime factorisation of 704 is as follows:

27042352217628824422211111

704 = 2 × 2 × 2 × 2 × 2 × 2 × 11

Here, one 11 is left which is not in a group of three.

If we divide 704 by 11, then it will become a perfect cube.

Thus, 704 ÷ 11 = 64 = 2 × 2 × 2 × 2 × 2 × 2 is a perfect cube.

Hence, the smallest number by which 704 should be divided to make it a perfect cube is 11.

Q.3 Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube?

Ans

Volume of the cube of sides 5 cm,2 cm ,5 cm

=5 cm ×5 cm ×2 cm =5 ×5 ×2 cm3

Here, two 5s and one 2 are left which are not in a triplet. If we multiply this expression by 5 × 2 × 2 = 20, then it will become a perfect cube.

Thus, 5 × 5 × 2 × 5 × 2 × 2 = 2 × 2 × 2 × 5 × 5 × 5 = 1000 is a perfect cube

Hence, 20 cuboids of 5 cm, 2 cm, and 5 cm are required to form a cube.

Q.4 Find the cube root of each of the following numbers by prime factorization method.

(i) 64 (ii) 512 (iii) 10648 (iv)27000 (v) 15625
(vi)13824 (vii) 110592 (viii) 46656 (ix) 175616
(x) 91125

Ans

(i) The prime factorisation of 64 is as follows:2642322162824221 64=2×2×2¯×2×2×2¯643=2×2=4 (ii) The prime factorisation of 512 is as follows:2512225621282642322162824221 512=2×2×2¯×2×2×2¯×2×2×2¯5123=2×2×2=8 (iii) The prime factorisation of 10648 is as follows:21064825324226621113311112111111 10648=2×2×2¯×11×11×11¯106483=2×11=22 (iv)The prime factorisation of 27000 is as follows:22700021350026750333753112533755125525551 27000=2×2×2¯×3×3×3¯×5×5×5¯270003=2×3×5=30 (v) The prime factorisation of 15625 is as follows:5156255312556255125525551 15625=5×5×5¯×5×5×5¯156253=5×5=25 (vi) The prime factorisation of 13824 is as follows:213824269122345621728286424322216210825432739331 13824=2×2×2¯×2×2×2¯×2×2×2¯×3×3×3¯138243=2×2×2×3=24 (vii) The prime factorisation of 110592 is as follows:2110592255296227648213824269122345621728286424322216210825432739331 110592=2×2×2¯×2×2×2¯×2×2×2¯×2×2×2¯×3×3×3¯1105923=2×2×2×2×3=48 (viii) The prime factorisation of 46656 is as follows:2466562233282116642583222916214583729324338132739331 46656=2×2×2¯×2×2×2¯×3×3×3¯×3×3×3¯466563=2×2×3×3=36 (ix) The prime factorisation of 175616 is as follows:217561628780824390422195221097625488227442137226867343749771 175616=2×2×2¯×2×2×2¯×2×2×2¯×7×7×7¯1756163=2×2×2×7=56 (x) The prime factorisation of 91125 is as follows:391125330375310125333753112533755125525551 91125=3×3×3¯×3×3×3¯×5×5×5¯911253=3×3×5=45

Q.5 State true or false.

(i) Cube of any odd number is even.

(ii) A perfect cube does not end with two zeros.

(iii) If square of a number ends with 5, then its cube ends with 25.

(iv)There is no perfect cube which ends with 8.

(v) The cube of a two digit number may be a three digit number.

(vi)The cube of a two digit number may have seven or more digits.

(vii) The cube of a single digit number may be a single digit number.

Ans

  1. False.

Reason: The cube of any odd number is an odd number because when we find the cube of any odd number then we are multiplying its unit’s digit three times and the unit’s digit of any odd number is also an odd number. Therefore, the product will again be an odd number.

For example: The cube of 9 (i.e., an odd number) is 729, which is again an odd number.

  1. True.

Reason: A perfect cube will end with a certain number of zeroes that are always a perfect multiple of 3.

For example: The cube of 10 is 1000 and there are 3 zeroes at the end of it.

The cube of 100 is 1000000 and there are 6 zeroes at the end of it.

  1. False.

Reason: It is not always necessary that if the square of a number ends with 5, then its cube will end with 25.

For example: the square of 25 is 625 and 625 has its unit digit as 5. The cube of 25 is 15625. However, the square of 35 is 1225 and also has its unit place digit as 5 but the cube of 35 is 42875 which does not end with 25.

  1. False.

Reason: The cubes of all the numbers having their unit’s digit as 2 will end with 8.

For example: The cube of 22 is 10648 and the cube of 32 is 32768.

  1. False.

Reason: The smallest two-digit natural number is 10, and the cube of 10 is 1000 which has 4 digits in it.

  1. False.

Reason: The largest two-digit natural number is 99, and the cube of 99 is 970299 which has 6 digits in it. Therefore, the cube of any two-digit number cannot have 7 or more digits in it.

(vii)True

Reason: As the cube of 1 and 2 are 1 and 8 respectively.

Q.6 You are told that 1,331 is a perfect cube. Can you guess without factorization what is its cube root? Similarly, guess the cube roots of 4913, 12167, 32768.

Ans

Firstly, we will make groups of three digits starting from the rightmost digit of the number 1331.

There are 2 groups, 1 and 331, in it.

Consider the first group 331.

The digit at its unit place is 1. We know that if the digit 1 is at the end of a perfect cube number, then its cube root will have its unit place digit as 1 only. Therefore, the unit place digit of the required cube root can be taken as 1.

Now we will take the second group i.e., 1.

The cube of 1 exactly matches with the number of the second group. Therefore, the tens digit of our cube root will be taken as the one’s place of the smaller number whose cube is near to the number of the second group i.e., 1 itself. 1 will be taken as tens place of the cube root of 1331.

Hence, 1331 3 =11 49133 MathType@MTEF@5@5@+=feaaguart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqabaWaaqaafaaakeaadaGcbaqaaiaaisdacaaI5aGaaGymaiaaiodaaSqaaiaaiodaaaaaaa@3CE0@

First make groups of three digits starting from the rightmost digit of 4913.

The groups are 4 and 913.

Consider the first group 913.

The number 913 ends with 3. We know that if a perfect cube number ends with 3, then its cube root will have its unit place digit as 7 only. Therefore, the unit place digit of the required cube root is taken as 7.

Now, take the second group i.e., 4.

We know that, 13 = 1 and 23 = 8

Also, 1 < 4 < 8

Therefore, we take the one’s place, of the smaller number 1 as the ten’s place of the required cube root.

Hence, 4913 3 =17

Now,

12167 3 MathType@MTEF@5@5@+=feaaguart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqabaWaaqaafaaakeaadaGcbaqaaiaaigdacaaIYaGaaGymaiaaiAdacaaI3aaaleaacaaIZaaaaaaa@3D9A@

First make groups of three digits starting from the rightmost digit of 12167.

The groups are 12 and 167.

Consider the first group 167.

167 end with 7. We know that if a perfect cube number ends with 7, then its cube root will have its unit’s digit as 3 only. Therefore, the unit place digit of the required cube root can be taken as 3.

Take the second group i.e., 12.

We know that, 23 = 8 and 33 = 27

Also, 8 < 12 < 27

2 is smaller between 2 and 3. Therefore, 2 will be taken at the tens place of the required cube root.

Hence, 12167 3 =23 MathType@MTEF@5@5@+=feaaguart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqabaWaaqaafaaakeaacaqGibGaaeyzaiaab6gacaqGJbGaaeyzaiaabYcacaqGGaWaaOqaaeaacaaIXaGaaGOmaiaaigdacaaI2aGaaG4naaWcbaGaaG4maaaakiabg2da9iaaikdacaaIZaaaaa@45E7@ 32768 3 MathType@MTEF@5@5@+=feaaguart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqabaWaaqaafaaakeaacaqGGaWaaOqaaeaacaaIZaGaaGOmaiaaiEdacaaI2aGaaGioaaWcbaGaaG4maaaaaaa@3E46@

We will make groups of three digits starting from the rightmost digit of the number 32768.

The groups are 32 and 768.

Consider the first group 768.

768 end with 8. We know that if a perfect cube number ends with 8, then its cube root will have its unit’s digit as 2 only. Therefore, the unit place digit of the required cube root will be taken as 2.

Taking the other group i.e., 32,

We know that, 33 = 27 and 43 = 64

Also, 27 < 32 < 64

3 is smaller between 3 and 4. Therefore, 3 will be taken at the tens place of the required cube root.

Hence, 32768 3 =32 MathType@MTEF@5@5@+=feaaguart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqabaWaaqaafaaakeaacaqGibGaaeyzaiaab6gacaqGJbGaaeyzaiaabYcacaqGGaWaaOqaaeaacaaIZaGaaGOmaiaaiEdacaaI2aGaaGioaaWcbaGaaG4maaaakiabg2da9iaaiodacaaIYaaaaa@45F0@

Q.7 Find the ratio of the following.

(a) Speed of a cycle 15 km per hour to the speed of scooter 30 km per hour.

(b) 5 m to 10 km

(c) 50 paise to ₹ 5

Ans

a Ratio of the speed of cycle to the speed of scooter=1530=1:2(b) Since 1 km = 1000 m,Required ratio=5m10 km=5m10×1000 m=1:2000(c) Since1 = 100 paise,Required ratio=50 paise₹ 5=50paise500paise=1:10

Q.8 Convert the following ratios to percentages.

(a) 3 : 4

(b) 2 : 3

Ans

(a) 3:4=34=34×100100=34×100%=75%(b) 2:3=23=23×100100=23×100%=2003%=6623%

Q.9 72% of 25 students are good in mathematics. How many are not good in mathematics?

Ans

Given that 72% of 25 students are good in mathematics.Therefore,Percentage of students who are not goodin mathematics =(100 72)%=28%.Thus,thenumber of students who are not good inmathematics=28100×25= 7Hence, 7 students are not good in mathematics.

Q.10 A football team won 10 matches out of the total number of matches they played. If their win percentage was 40, then how many matches did they play in all?

Ans

Let the total number of matches played by the team be x.It is given that the team won 10 matches and the winningpercentage of the team was 40%.Therefore,40100×x=10x=10×10040x=25Hence, the team played 25 matches in all.

Q.11 If Chameli had ₹ 600 left after spending 75% of her money, how much did she have in the beginning?

Ans

Let the amount of money which Chameli had in thebeginning be x.It is given that after spending 75% of ₹ x, shewas left with ₹ 600.Therefore,(100 75)% of x=Rs 600 25% of x=60025100×x=600x=(600×10025)= 2400Hence, she had ₹ 2400 in the beginning.

Q.12 If 60% people in a city like cricket, 30% like football and the remaining like other games, then what per cent of the people like other games? If the total number of people are 50 lakh, find the exact number who like each type of game.

Ans

Percentage of people who like other games=(1006030)% =(10090)% =10%It is given that the total number of people=50 lakhNumber of people who like cricket=(60100×50)lakh=30lakhNumber of people who like football=(30100×50)lakh=15lakhNumber of people who like other games=(10100×50)lakh=5lakh

Q.13 A man got a 10% increase in his salary. If his new salary is ₹ 1,54,000, find his original salary.

Ans

Let the original salary be x. It is given that the new salary is ₹ 1,54,000.Therefore, Original salary + Increment = New salaryBut it is given that the increment is 10% of the original salary.Therefore,we havex+10100×x=1,54,000110x100=1,54,000x=1,54,000×100110x=1,40,000Thus, the original salary was ₹ 1,40,000.

Q.14 A shopkeeper buys 80 articles for ₹ 2,400 and sells them for a profit of 16%. Find the selling price of one article.

Ans

It is given that the shopkeeper buys 80 articles for 2,400.Cost Price of one article=2,40080=30Profit percent=16%Profit percent=ProfitCost Price×10016=Profit30×100Profit=16×30100= 4.80 Selling price of one article=C.P. + Profit=(30+4.80)= 34.80

Q.15 The cost of an article was ₹ 15,500. ₹ 450 were spent on its repairs. If it is sold for a profit of 15%, find the selling price of the article.

Ans

The cost of an article was ₹ 15,500
The amount spent on its repairs ₹ 450

Total cost of an article=Cost+Overhead expenses=₹ 15500+Rs 450=₹ 15950Profit%=ProfitC.P.×100 15=Profit15,950×100Profit= (15,950×15100) = 2392.50We know, S.P. of an article=C.P.+Profit=(15950+2392.50)=₹ 18,342.50

Q.16 A VCR and TV were bought for ₹ 8,000 each. The shopkeeper made a loss of 4% on the VCR and a profit of 8% on the TV. Find the gain or loss percent on the whole transaction.

Ans

C.P. of a VCR = ₹ 8000 Loss% = 4 % Let us suppose that C.P. is ₹ 100Since loss is 4% ,so S.P. is ₹ 96.When C.P. is Rs 8000, S.P.=(96100×8000)= 7680Now, C.P. of a TV = ₹ 8000The shopkeeper made a profit of 8 % on TV.Let us suppose that C.P. of TV is ₹ 100Then S.P. is ₹ 108.When C.P. is Rs 8000, S.P.=(108100×8000)= 8640To find the profit or loss on whole transaction,we have to find total C.P.and S.P.Total S.P.= 7680+ 8640= 16320Total C.P.= 8000+ 8000= 16000Since total S.P. is greater than total C.P.,therefore,we have a profit. Profit =Total S.PTotal C.P. = 16320 16000 = 320Now,Profit%=ProfitC.P.×100 =32016000×100 =2%Hence, the shopkeeper had a gain of 2% on the whole transaction.

Q.17 During a sale, a shop offered a discount of 10% on the marked prices of all the items. What would a customer have to pay for a pair of jeans marked at ₹ 1450 and two shirts marked at ₹ 850 each?

Ans

Total marked price= (1,450+2×850) = (1,450+1,700) = 3,150Given that, discount %=10%Discount= (3,150×10100)= 315Also, Discount=Marked price Sale price ₹ 315= 3150Sale price Sale price= (3150315) = 2835Thus, the customer will have to pay ₹ 2,835.

Q.18 A milkman sold two of his buffaloes for ₹ 20,000 each.On one he made a gain of 5% and on the other a loss of 10%. Find his overall gain or loss. (Hint: Find CP of each)

Ans

Given,S.P. of each buffalo=20000 Gain%=5% This means if C.P. is100, then S.P. is105.C.P. of one buffalo= 20000×100105= 19047.62Also,the second buffalo was sold at a loss of 10%.This means if C.P. is100, then S.P. is90.C.P. of other buffalo= 20000×10090= 22222.22Total C.P.= 19047.62+ 22222.22= 41269.84 Total S.P.=20000 +20000=40000Loss= 41269.8440000= 1269.84Thus, the overall loss of the milkman was1,269.84.

Q.19 The price of a TV is ₹ 13,000. The sales tax charged on it is at the rate of 12%. Find the amount that Vinod will have to pay if he buys it.

Ans

The price of the TV = ₹ 13,000Sales tax =12%i.e.= ₹(12100×13,000) = 1,560Required amount=Cost+Sales Tax=₹ 13000+₹ 1560= 14560Therefore, Vinod have to pay ₹ 14,560 for the T.V.

Q.20 Arun bought a pair of skates at a sale where the discount given was 20%. If the amount he pays is ₹ 1,600, find the marked price.

Ans

Let the marked price be x.Discount%=DiscountMarked price×10020=Discountx×100Discount=20x100=15xAlso,Discount = Marked priceSale price15x=x 1600x15x= 160045x= 1600x=1600×54= 2000The marked price of the scates was ₹ 2000.

Q.21 I purchased a hair-dryer for ₹ 5,400 including 8% VAT. Find the price before VAT was added.

Ans

It is given that the price of the hair drier includes VAT.Let us suppose the price without VAT be ₹ 100, thenthe price including VAT will be ₹ 108.When price including VAT is ₹ 108, original price= 100When price including VAT is ₹ 5400,original price=(100108×5400)= 5000Thus, the price of the hairdryer before the addition of VAT was ₹ 5,000.

Q.22 

Calculate the amount and compound interest ona ₹ 10,800 for 3yearsat1212% per annumcompoundedannually.(b) ₹ 18,000 for 212yearsat10%per annumcompoundedannually.(c) ₹ 62,500 for 112yearsat8%per annumcompoundedhalfyearly.(d) ₹ 8,000 for 1yearat9%per annumcompoundedhalfyearly. (YoucouldusetheyearbyyearcalculationusingSIformulatoverify).(e) ₹ 10,000 for 1yearat8%per annumcompoundedhalfyearly.

Ans

(a) Principal= 10,800 R=1212%=252% n=3Amount=P(1+R100)n =10800(1+252×100)3 =10800(1+25200)3 =10800(225200)3 =10800×225200×225200×225200 =15377.34C.I=AmountPrincipal =15377.3410800 =4,577.34 (b) Principal= 18,000 R=10% n=212Amount=P(1+R100)n =18000(1+10100)212Since,the years are in fractional form,so we will calculatethe Amount for the whole part first and then we will calculatethe simple interest for the remaining fractional part.Amount=18000(1+10100)2 =18000(1110)2 =18000×1110×1110 = 21,780 Now,we will take ₹ 21,780 as principal and calculate the simple interest for the next 12year.S.I=PRT100 =21780×10×12×100 = 1089 Interest for the first 2 years=(2178018000)= 3780And interest for the next 12 year= 1089 Total C.I.= 3780+ 1089= 4,869, andA = P + C.I.= 18000+ 4869= 22,869 (c) Principal= 62,500 R=8% yearly=4% half yearly n=112=3 half yearsAmount=P(1+R100)n =62500(1+4100)3 =62500(2625)3 =62500×2625×2625×2625 = 70,304C.I=AP= 70,304 62,500 = 7,804(d) Principal= 8000 R=9% yearly=92% half yearly n=1=2 half yearsAmount=P(1+R100)n =8000(1+9200)2 =8000(209200)2 = 8,736.20C.I=AP= 8,736.20Rs 8000 = 736.20(e) Principal=Rs 10,000 R=8% yearly=4% half yearly n=1=2 half yearsAmount=P(1+R100)n =10,000(1+4100)2 =10,000(1+125)2 =10,000(2625)2 =10,000×2625×2625 = 10,816C.I=AP=Rs 10,816Rs 10,000 = 816

Q.23 Kamala borrowed ₹ 26,400 from a Bank to buy a scooter at a rate of 15% p.a. compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan?
(Hint: Find A for 2 years with interest is compounded yearly and then find SI on the 2nd year amount for 4/12 years).

Ans

Principal= 26,400 R=15% per annum n=2412yearsSince,the years are in fractional form,so we will calculatethe Amount for the whole part first and then we will calculatethe simple interest for the remaining fractional part.Amount=26,400(1+15100)2 =26,400(2320)2 =26,400×2320×2320 = 34,914Now,we will take Rs 34,914 as principal and calculate the simple interest for the next 13years.S.I=PRT100 =34,914×15×13×100 = 1745.70 Interest for the first 2 years=(34,91426,400)= 8,514And interest for the next 13year= 1745.70 Total C.I.= 8,514+ 1745.70= 10,259.70, andA = P + C.I.= 26,400+ 10,259.70= 36,659.70

Q.24 Fabina borrows ₹ 12,500 at 12% per annum for 3 years at simple interest and Radha borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much?

Ans

Interest paid by Fabina=PRT100 =(12,500×12×3100) = 4,500Amount paid by Radha=P(1+R100)n = 12,500(1+10100)3 = 12,500×(110100)3 = 16,637.50C.I=AP= 16,637.50 12,500 =4,137.50The interest paid by Fabina is4,500 and by Radha is4,137.50.Thus, Fabina pays more interest.i.e.4500 4137.50= 362.50Hence, Fabina will have to pay362.50 more.

Q.25 I borrowed ₹ 12,000 from Jamshed at 6% per annum simple interest for 2 years. Had I borrowed this sum at 6% per annum compound interest, what extra amount would I have to pay?

Ans

Principal= 12,000 R=6% yearly n=2 yearsSimple Interest =PRT100 =(12,000×6×2100) = 1,440Amount=P(1+R100)n =12,000(1+6100)2 =12,000(1+350)2 =12,000(5350)2 =12,000×5350×5350 = 13,483.20C.I=AP= 13,483.20 12,000= 1,483.20 C.IS.I= 1,483.20 1,440= 43.20Therefore, the extra amount to be paid is43.20.

Q.26 Vasudevan invested ₹ 60,000 at an interest rate of 12% per annum compounded half yearly. What amount would he get

(i) after 6 months?

(ii) after 1 year?

Ans

(i) Principal= 60,000 R=12% yearly=6% half yearly n=6 months=1 half yearAmount=P(1+R100)n =60,000(1+6100)1 =60,000(106100) = 63,600(ii) Principal= 60,000 R=12% yearly=6% half yearly n=1 year=2 half yearsAmount=P(1+R100)n =60,000(1+6100)2 =60,000(106100)2 = 60,000×106100×106100 = 67,416

Q.27 Arif took a loan of ₹ 80,000 from a bank. If the rate of interest is 10% per annum, find the difference in amounts he would be paying after

1 1 2 MathType@MTEF@5@5@+=feaaguart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiGacaGaaeqabaWaaqaafaaakeaacaaIXaWaaSaaaeaacaaIXaaabaGaaGOmaaaaaaa@3B51@

years if the interest is

(i) compounded annually.

(ii) compounded half yearly.

Ans

(i) Principal= 80,000 R=10% per annum n=112yearsSince,the years are in fractional form,so we will calculatethe Amount for the whole part first and then we will calculatethe simple interest for the remaining fractional part.Amount=80,0001+101001 =80,0001+110 =80,00011102 = 88,000Now,we will take Rs 88,000 as principal and calculate the simple interest for the next 12year.S.I=PRT100 =88,000×10×12×100 = 4,400Interest for the first year = ₹ 88000 ₹ 80000 = ₹ 8,000And interest for the next 12year = ₹ 4,400Total C.I. = ₹ 8000 + ₹ 4,400= ₹ 1,2400A = P + C.I. =80000 + 12400 = ₹ 92,400 ii Now,the interest is compounded half yearly.Rate=10% per annum=5% per half yearn= 112years=3 half years Amount=80,0001+51003 =80,0001+1203 =80,00021202 = 92,610

Therefore, the difference between the amounts = ₹ 92,610 − ₹ 92,400 = ₹ 210

Q.28 Maria invested ₹ 8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find

(i) The amount credited against her name at the end of the second year.

(ii) The interest for the 3rd year.

Ans

(i) Principal= 8,000 R=5% yearly n=2 yearsAmount=P(1+R100)n =8,000(1+5100)2 =8,000(2120)2 = 8,000×2120×2120 = 8,820 (ii) Now,we have to calculate the interest for the third year. So we will take Rs 8,820 as principal and calculate the simple interest for the next year.S.I=PRT100 =8820×5×1100 = 441

Q.29 Find the amount and the compound interest on ₹10,000 for years at 10% per annum, compounded half yearly. Would this interest be more than the interest he would get if it was compounded annually?

Ans

Principal= 10,000 R=10% yearly=5% half yearly n=112 years=3 half yearsAmount=P(1+R100)n =10,000(1+5100)3 =10,000(1+120)3 =10,000(2120)3 = 60,000×2120×2120×2120 = 11,576.25C.I=AP = 11,576.25Rs 10,000 = 1,576.25Now we have to calculate the interest compounded annually.Since,the years are in fractional form,so we will calculatethe Amount for the whole part first and then we will calculatethe simple interest for the remaining fractional part.Amount= 10,000(1+10100)1 = 10,000(1110) =11,000 Now,we will take 11,000 as principal and calculate the simple interest for the next 12years.S.I=PRT100 =11,000×10×12×100 = 550 Interest for the first year=11000 10000= 1,000 Total compound interest=1000+ 550= 1,550Yes,the interest would be more when compounded half yearlythan the interest when compounded annually.

Q.30 Find the amount which Ram will get on ₹ 4096, if he gave it for 18 months at

1212

% 2 per annum, interest being compounded half yearly.

Ans

Principal= 4,096 R=1212% yearly=254% half yearly n=18 months=3 half yearsAmount=P(1+R100)n = 4096(1+25400)3 = 4096(1+116)3 = 4096(1716)3 =4096×1716×1716×1716 =4,913Therefore, the required amount is ₹ 4,913.

Q.31 The population of a place increased to 54,000 in 2003 at a rate of 5% per annum

(i) find the population in 2001.

(ii) what would be its population in 2005?

Ans

(i) The population in the year 2003= 54,000 54,000=(Population in 2001)(1+5100)2 54,000=(Population in 2001)(2120)2 Population in 2001=54,000×(2021)2 =48979.59Therefore, the population in the year 2001 was approximately 48,980.(ii) Population in 2005=54,000×(1+5100)2 =54,000×(1+120)2 =54,000×(2120)2 =59,535Therefore,the population in the year 2005 would be 59,535.

Q.32 In a Laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially 5, 06,000.

Ans

The initial count of bacteria is given as 5,06,000. Bacteria at the end of 2 hours =5,06,000 ( 1+ 1 40 ) 2 =5,06,000 ( 41 40 ) 2 =5,06,000× 41 40 × 41 40 = 531616.25 Thus, the count of bacteria at the end of 2 hours will be 5,31,616 ( approx. ). MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@F34A@

Q.33 A scooter was bought at ₹ 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.

Ans

Principal=Cost price of the scooter= 42,000Depreciation=8%of42,000 = 42000×8100= 3,360Value after 1 year= 42000 3360= 38,640.

Q.34 Add the following.
(i) ab – bc, bc – ca, ca – ab
(ii) a – b + ab, b – c + bc, c – a + ac
(iii) 2p2q2 – 3pq + 4, 5 + 7pq – 3p2q2
(iv) l2 + m2, m2 + n2, n2 + l2,2lm + 2mn + 2nl

Ans

(i)abbc bcca + ab +ca¯ 0(ii) ab+ab b c+bc + a +c +ac¯ ab + bc + ac(iii) 2p2q23pq+4+ 3p2q2+7pq+5¯ –1p2q2+4pq+9 iv l2+m2+ m2+n2l2+n2 +2lm+2mn+2nl¯ 2l2+2m2+2n2+2lm+2mn+2nl

Q.35 (a) Subtract 4a – 7ab + 3b + 12 from 12a – 9ab + 5b – 3

(b) Subtract 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + 10xyz

(c) Subtract 4p2q – 3pq + 5pq2 – 8p + 7q – 10 from 18 – 3p – 11q + 5pq – 2pq2 + 5p2q

Ans

(a) 12a 9ab+ 5b 3 4a 7ab+ 3b+ 12(–) (+) (–) (–)¯ 8a 2ab+ 2b 15¯ (b) 5xy 2yz 2zx+ 10xyz 3xy+ 5yz 7zx(–) (–) (+)¯ 2xy – 7yz+5zx+10xyz¯(c) 18 3p 11q+ 5pq 2pq2+ 5p2q 10 8p + 7q 3pq + 5pq2+4p2q (+) (+) () (+) () ()¯ 28 +5p 18q +8pq 7pq2 +1p2q¯

Q.36 Find the product of the following pairs of monomials.

(I) 4, 7p

(ii) – 4p, 7p

(iii) – 4p, 7pq

(iv) 4p3, –3p

(v) 4p, 0

Ans

(i) 4 × 7p = 4 × 7 × p = 28p

(ii) –4p × 7p = (–4) × 7 × p × p = –28p2

(iii) –4p × 7pq =(–4) × 7 × p × p × q = –28p2q

(iv) 4p3 × (–3p) = 4 × (–3) × p3 × p = –12p4

(v) 4p × 0 = 4 × p × 0 = 0

Q.37 Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.

(p, q); (10m, 5n); (20x2, 5y2); (4x, 3x2); (3mn, 4np)

Ans

          Area of rectangle = Length × BreadthArea of first rectangle = p×q =pq Area of second rectangle = 10m× 5n =10×5×m×n =50 mn Area of third rectangle =20x2×5y2 =20×5×x2×y2 =100 x2y2Area of fourth rectangle =4x×3x2 =4×3×x×x2 = 12x3 Area of fifth rectangle =3mn×4np =3×4×m×n×n×p =12 mn2p

Q.38 Complete the table of products.

FirstmonomialSecondmonomial2x5y3x24xy7x2y9x2y22x4x2...............5y......15x2y.........3x2..................4xy..................7x2y..................9x2y2..................

Ans

FirstmonomialSecondmonomial2x5y3x24xy7x2y9x2y22x4x210xy¯6x3¯8x2y¯14x3y¯18x3y2¯5y10xy¯25y2¯15x2y20xy2¯35x2y2¯45x2y3¯3x26x3¯15x2y¯9x4¯12x3y¯21x4y¯27x4y2¯4xy8x2y¯20xy2¯12x3y¯16x2y2¯28x3y2¯36x3y3¯7x2y14x3y¯35x2y2¯21x4y¯28x3y2¯49x4y2¯63x4y3¯9x2y218x3y2¯45x2y3¯27x4y2¯36x3y3¯63x4y3¯81x4y4¯

Q.39 Obtain the volume of rectangular boxes with the following length, breadth and height respectively.

(I) 5a, 3a2, 7a4

(ii) 2p, 4q, 8r

(iii) xy, 2x2y, 2xy2

(iv) a, 2b, 3c

Ans

2 × 7a4

2 × a4

7

xy × 2x2y × 2xy2

x × x2 × x × y × y × y2

x4y4

a × 2b × 3c

× a × b × c

Q.40 Obtain the product of

(i) xy, yz, zx

(ii) a, – a2, a3

(iii) 2, 4y, 8y2, 16y3

(iv) a, 2b, 3c, 6abc

(v) m, – mn, mnp

Ans

(i) xy×yz×zx=x×y×y×z×z×x=x2y2z2 (ii) a×(a2)×a3=a6(iii) 2×4y×8y2×16y3=2×4×8×16 ×y×y2×y3=1024y6(iv) a×2b×3c×6abc=2×3×6×a×b×c×abc=36a2b2c2 (v) m×(mn)×mnp=m×(m)×n×m×n×p=m3n2p

Q.41 Carry out the multiplication of the expressions in each of the following pairs.
(i) 4p, q + r
(ii) ab, a – b
(iii) a + b, 7a2b2
(iv) a2 – 9, 4a
(v) pq + qr + rp, 0

Ans

(i) 4p×(q+r)=(4p×q)+(4p×r)=4pq+4pr(ii) ab×(ab)=ab×aab×b=a2bab2 (iii)(a+b)× 7a2b2= a ×7a2b2+b×7a2b2= 7a3b2 + 7a2b3 (iv)a2 9× 4a=a2× 4a 9×4a=4a336a (v) (pq+qr+rp)× 0=(pq×0)+(qr×0)+(rp ×0)= 0

Q.42 Complete the table.

First ExpressionSecond ExpressionProduct(i)ab+c+d...(ii)x+y55xy...(iii)p6p27p+5...(iv)4p2q2p2q2...(v)a+b+cabc...

Ans

First ExpressionSecond ExpressionProduct(i)ab+c+dab+ac+ad(ii)x+y55xy5x2y+5xy225xy(iii)p6p27p+56p37p2+5p(iv)4p2q2p2q24p4q24p2q4(v)a+b+cabca2bc+ab2c+abc2

Q.43

Findtheproduct.i(a2)×(2a22)×(4a26)ii23xy×910x2y2iii103pq3×65p3qivx×x2×x3×x4

Ans

i(a2)×(2a22)×(4a26)=2×4×a2×a22×a26=8a50ii(23xy)×(910x2y2)=23×910×x×x2×y×y2=35x3y3iii(103pq3)×(65p3q)=103×65×p×q3×p3×q=4p4q4ivx×x2×x3×x4=x10

Q.44 

(a) Simplify 3x(4x5)+3 and find its values fori x=3 ii x=12(b) Simplify a(a2+a+1)+5 and find its value for(i) a=0, (ii) a=1, (iii) a=1

Ans

(a)3x (4x – 5) + 3= 3x×4x-3x×5+3= 12x2-15x+3(i) For x = 3, 12x2-15x+3= 1232-15×3+3= 108-45+3= 66 ii For x=12,12x215x+3=12(12)215×12+3=12×14152+3=3152+3=6152=12152=32(b)a(a2+a+1)+5=a×a2+a×a+a×1+5=a3+a2+a+5(i)For a=0,a3+a2+a+5=0+0+0+5=5(ii) For a=1,a3+a2+a+5=13+12+1+5=8iii For a=1,a3+a2+a+5=(1)3+(1)2+(1)+5=(1)+1+(1)+5=4

Q.45 

(a) Add:p(pq),q(qr) and r(rp)(b) Add:2x(zxy) and 2y(zyx)(c) Subtract : 3l(l4m+5n) from 4l(10n3m+2l)(d) Subtract : 3a(a+b+c)2b(ab+c)from 4c(a+b+c)

Ans

First we will simplify the given expressions and then add or subtract them.(a)p(pq)=p2pqq(qr)=q2qrr(rp)=r2rpNow,adding all the expressions p2pqq2qr + r2rp ¯ p2pq+q2qr+r2rp           ¯(b) 2x(zxy)=2xz2x22xy2y(zyx)=2yz2y22yxNow,adding all the expressions 2xz2x22xy+ 2yx+2yz2y2¯ 2xz2x24xy+2yz2y2    ¯         (c) 3l(l4m+5n)=3l212lm+15ln4l(10n3m+2l)=40ln12lm+8l2 On subtracting them, we get 40ln12lm+8l2 15ln12lm+3l2()       (+) ()¯ 25ln +5l2 (d) 3a(a+b+c)2b(ab+c)=3a2+ab+3ac+2b22bc4c(a+b+c)=4ac+4cb+4c2 On subtracting them, we get 4ac+4cb+4c2 +3ac2bc +3a2+ab+2b2 () (+) ()​ () ()¯ 7ac + 6bc + 4c23a2ab2b2¯

Q.46 

Multiplythebinomials.(i)(2x+5)and(4x3)(ii)(y8)and(3y4)(iii)(2.5l0.5m)and(2.5l+0.5m)(iv)(a+3b)and(x+5)v(2pq+3q2)and(3pq2q2)vi34a2+3b2and4(a223b2)

Ans

(i)(2x+5) and (4x3)=2x×(4x3)+5×(4x3)=8x26x+20x15=8x2+14x15(ii)(y8) and (3y4)=y×(3y4)8×(3y4)=3y24y24y+32=3y2+3228y(iii)(2.5l0.5m) and (2.5l + 0.5m)=2.5l×(2.5l+0.5m)0.5m×(2.5l+0.5m)=6.25l2+1.25lm1.25lm0.25m2=6.25l20.25m2(iv)(a+3b) and (x+5)=a×(x+5)+3b×(x+5)=ax+5a+3bx+15b(v)(2pq+3q2) and (3pq2q2)=2pq×(3pq2q2)+3q2×(3pq2q2)=6p2q24pq3+9pq36q4=6p2q2+5pq36q4vi34a2+3b2 and 4(a223b2)=(34a2)×4(a223b2)+(3b2)×4(a223b2)=3a42a2b2+12a2b28b4=3a4+10a2b28b4

Q.47 Find the product.

(i) (5 – 2x) (3 + x)

(ii) (x + 7y) (7x – y)

(iii) (a2 + b) (a + b2)

(iv) (p2 – q2) (2p + q)

Ans

i 5 2x 3 +x=5×3 +x2x3 +x                          =15+5x6x2x2                          =15x2x2ii x+ 7y 7xy=x×7xy+7y×7xy                           =7x2xy+49xy7y2                           =7x2+48xy7y2iii a2+b a+b2=a2×a+b2+b×a+b2                          =a3+a2b2+ba+b3iv p2q2 2p+q=p2×2p+qq2×2p+q                           =2p3+p2q2q3q3                           =2p3+p2q3q3

Q.48 Simplify.

(i) (x2 – 5) (x + 5) + 25

(ii) (a2 + 5) (b3 + 3) + 5

(iii) (t + s2) (t2 – s)

(iv) (a + b) (c – d) + (a – b) (c + d) + 2 (ac + bd)

(v) (x + y)(2x + y) + (x + 2y)(x – y)

(vi) (x + y)(x2 – xy + y2)

(vii) (1.5x – 4y)(1.5x + 4y + 3) – 4.5x + 12y

(viii) (a + b + c)(a + b – c)

Ans

(i) (x2 5)(x+ 5) + 25= x2(x+ 5)5 (x+ 5)+25=x3+5x25x25+25=x3+5x25x(ii) (a2+ 5)(b3+ 3) + 5=a2(b3+ 3)+5(b3+ 3)+5=a2b3+3a2+5b3+15+5=a2b3+3a2+5b3+20(iii) (t+s2)(t2s)=t(t2s)+s2(t2s)=t3ts+s2t2s3(iv)(a+b)(cd) + (ab)(c+d)+2 (ac+bd)=a(cd)+b(cd)+a(c+d)b(c+d)+2 (ac+bd)=acad+bcbd+ac+adbcbd+2ac+2bd=4ac(v) (x+y)(2x+y) + (x+ 2y)(xy) =x(2x+y)+y(2x+y)+x(xy) +2y(xy) =2x2+xy+2xy+y2+x2xy+2xy2y2=3x2y2+4xy (vi) (x+y)(x2xy+y2)=x(x2xy+y2)+y(x2xy+y2)=x3x2y+xy2+yx2xy2+y3=x3+y3(vii) (1.5x 4y)(1.5x+ 4y+ 3) 4.5x+ 12y=1.5x(1.5x+ 4y+ 3)4y(1.5x+ 4y+ 3) 4.5x+ 12y=2.25x2+6xy+4.5x6xy16y212y4.5x+12y=2.25x216y2(viii) (a+b+c)(a+bc) =a(a+bc) +b(a+bc)+c (a+bc) =a2+abac+ab+b2bc+ac+cbc2=a2+b2c2+2ab

Q.49 

Useasuitableidentitytogeteachofthefollowingproducts.(i)(x+3)(x+3)(ii)(2y+5)(2y+5)iii2a72a7iv3a123a12(v)(1.1m0.4)(1.1m+0.4)(vi)(a2+b2)(a2+b2)(vii)(6x7)(6x+7)(viii)(a+c)(a+c)ixx2+3y4x2+3y4x7a9b7a9b

Ans

ix+3x+3=(x+3)2=(x)2+2×3×x+(3)2 {By using the identity(a+b)2=a2+b2+2ab}=x2+6x+9ii2y+52y+5=(2y+5)2=(2y)2+2×5×2y+(5)2 {By using the identity(a+b)2=a2+b2+2ab}=4y2+20y+25iii2a72a7=(2a7)2=(2a)22×2a×7+(7)2 {By using the identity(ab)2=a2+b22ab}=4a228a+49iv3a123a12=(3a12)2=(3a)22×3a×12+(12)2 {By using the identity(ab)2=a2+b22ab}=9a23a+14(v)(1.1m0.4)(1.1m+0.4)=(1.1m)2(0.4)2 {By using the identity(a+b)(ab)=a2b2}=1.21m20.16(vi)(a2+b2)(a2+b2)=(b2)2(a2)2 {By using the identity(a+b)(ab)=a2b2}=b4a4(vii)(6x7)(6x+7)=(6x)2(7)2 {By using the identity(a+b)(ab)=a2b2}=36x249(viii)(a+c)(a+c)=(a+c)2 {By using the identity(a+b)2=a2+b2+2ab} =c2+a22ac ix (x2+3y4)(x2+3y4)=(x2+3y4)2=(x2)2+(3y4)2+2(x2)(3y4) {By using the identity(a + b)2=a2+b2+2ab}= x24+9y216+3xy4x 7a9b7a9b=(7a9b)2=(7a)2+(9b)22(7a)(9b) {By using the identity(ab)2=a2+b22ab}=49a2+81b2126ab

Q.50 

Usetheidentity(x+a)(x+b)=x2+(a+b)x+abtofindthefollowingproducts.(i)(x+3)(x+7)(ii)(4x+5)(4x+1)(iii)(4x5)(4x1)(iv)(4x+5)(4x1)(v)(2x+5y)(2x+3y)(vi)(2a2+9)(2a2+5)(vii)(xyz4)(xyz2)

Ans

i(x+3)(x+7)=x2+(3+7)x+3×7=x2+10x+21ii4x+54x+1=(4x)2+5+1(4x)+5×1=16x2+24x+5iii4x54x1=(4x)2+[(5)+(1)](4x)+(5)×(1)=16x224x+5iv4x+54x1=(4x)2+[5+(1)](4x)+5×(1)=16x2+16x5v2x+5y2x+3y=(2x)2+5y+3y(2x)+5y×3y=4x2+16xy+15y2vi2a2+92a2+5=(2a2)2+9+5(2a2)+9×5=4a4+28a2+45viixyz4xyz2=(xyz)2+[(4)+(2)](xyz)+(4)×(2)=x2y2z26xyz+8

Q.51 

Findthefollowingsquaresbyusingtheidentities.i(b7)2ii(xy+3z)2iii(6x25y)2iv(23m+32n)2v(0.4p0.5q)2vi(2xy+5y)2

Ans

i(b7)2=(b)22×b×7+(7)2 {By using the identity(a b)2=a2+b22ab}=b214b+49ii(xy+3z)2=(xy)2+2×xy×3z+(3z)2 {By using the identity(a + b)2=a2+b2+2ab}=x2y2+6xyz+9z2iii(6x25y)2=(6x2)22×6x2×5y+(5y)2 {By using the identity(a b)2=a2+b22ab}=36x460x2y+25y2iv(23m+32n)2=(23m)2+2×23m×32n+(32n)2{By using the identity(a + b)2=a2+b2+2ab}=49m2+2mn+94n2v(0.4p0.5q)2=(0.4p)22×0.4p×0.5q+(0.5q)2 {By using the identity(a b)2=a2+b22ab}=0.16p20.4pq+0.25q2vi(2xy+5y)2=(2xy)2+2×2xy×5y+(5y)2 {By using the identity(a + b)2=a2+b2+2ab}=4x2y2+20xy2+25y2

Q.52 

Simplify.i(a2b2)2ii(2x+5)2(2x5)2iii(7m8n)2+(7m+8n)2iv(4m+5n)2+(5m+4n)2v(2.5p1.5q)2(1.5p2.5q)2vi(ab+bc)22ab2cvii(m2n2m)2+2m3n2

Ans

i(a2b2)2=(a2)22×a2×b2+(b2)2{By using the identity(a b)2=a2+b22ab}=(a)42a2b2+(b)4ii(2x+5)2(2x5)2=[(2x)2+2×2x×5+(5)2][(2x)22×2x×5+(5)2] {By using the identity(a + b)2=a2+b2+2ab}{By using the identity(ab)2=a2+b22ab}=4x2+20x+254x2+20x25=40xiii(7m8n)2+(7m+8n)2=[(7m)22×7m×8n+(8n)2]+[(7m)2+2×7m×8n+(8n)2] {By using the identity(a b)2=a2+b22ab}{By using the identity(a + b)2=a2+b2+2ab}=49m2112mn+64n2+49m2+112mn+64n2=98m2+128n2iv(4m+5n)2+(5m+4n)2=[(4m)2+2×4m×5n+(5n)2]+[(5m)2+2×5m×4n+(4n)2] {By using the identity(a + b)2=a2+b2+2ab}=(16m2+40mn+25n2)+(25m2+40mn+16n2)=41m2+80mn+41n2v(2.5p1.5q)2(1.5p2.5q)2=[(2.5p)22×2.5p×1.5q+(1.5q)2][(1.5p)22×1.5p×2.5q+(2.5q)2] {By using the identity(a b)2=a2+b22ab}=6.25p27.5pq+2.25q2[2.25p27.5pq+6.25q2]=6.25p27.5pq+2.25q22.25p2+7.5pq6.25q2=4p24q2 vi(ab+bc)22ab2c=[(ab)2+2×ab×bc+(bc)2]2ab2c {By using the identity(a + b)2=a2+b2+2ab}=a2b2+2ab2c+b2c22ab2c=a2b2+b2c2vii(m2n2m)2+2m3n2=[(m2)22×m2×n2m+(n2m)2]+2m3n2 {By using the identity(a b)2=a2+b22ab}=m42m3n2+n4m2++2m3n2=m4+n4m2

Q.53 

Showthat.i (3x+7)284x=(3x7)2ii (9p5q)2+180pq=(9p+5q)2iii (43m34n)2+2mn=169m2+916n2iv (4pq+3q)2(4pq3q)2=48pq2v (ab)(a+b)+(bc)(b+c)+(ca)(c+a)=0

Ans

i L.H.S=(3x+7)284x=(3x)2+72+2×3x×784x=9x2+49+42x84x=9x2+4942x R.H.S=(3x7)2=(3x)2+722×3x×7=9x2+4942xL.H.S=R.H.SHence ProvediiL.H.S=(9p5q)2+180pq=(9p)2+(5q)22×9p×5q+180pq=81p2+25q290pq+180pq=81p2+25q2+90pq R.H.S=(9p+5q)2=(9p)2+(5q)2+2×9p×5q=81p2+25q2+90pqL.H.S=R.H.SHence ProvediiiL.H.S=(43m34n)2+2mn=(43m)2+(34n)22×43m×(34n)+2mn=169m2+916n22mn+2mn=169m2+916n2 R.H.S=169m2+916n2L.H.S=R.H.SHence Proved iv L.H.S=(4pq+3q)2(4pq3q)2=48pq2=(4pq)2+(3q)2+2×4pq×(3q)[(4pq)2+(3q)22×m4pq×(3q)]=16p2q2+9q2+24pq2[16p2q2+9q224pq2]=16p2q2+9q2+24pq216p2q29q2+24pq2=48pq2 R.H.S=48pq2L.H.S=R.H.SHence Proved(v)L.H.S=(ab)(a+b)+(bc)(b+c)+(ca)(c+a)=a2b2+b2c2+c2a2=0 R.H.S=0L.H.S=R.H.SHence Proved

Q.54 

Usingidentities,evaluate.(i)712(ii)992(iii)1022(iv)9982(v)5.22(vi)297×303(vii)78×82(viii)8.92(ix)10.5×9.5

Ans

i712=(70+1)2=(70)2+2×70×1+12 {By using the identity(a + b)2=a2+b2+2ab}=4900+140+1=5041ii992=(1001)2=(100)22×100×1+12 {By using the identity(a b)2=a2+b22ab}=10000200+1=9801iii1022=(100+2)2=(100)2+2×100×2+22 {By using the identity(a + b)2=a2+b2+2ab}=10000+400+4=10404iv9982=(10002)2=(1000)22×1000×2+22 {By using the identity(a b)2=a2+b22ab}=10000004000+4=996004 (v)5.22=(5+0.2)2=(5)2+2×5×0.2+(0.2)2 {By using the identity(a + b)2=a2+b2+2ab}=25+2+0.04=27.04(vi)297×303=(3003)(300+3)=(300)232 {By using the identity(ab)(a+b)=a2b2}=900009=89991(vii)78×82=(802)(80+2)=(80)222 {By using the identity(ab)(a+b)=a2b2}=64004=6396viii8.92=(9.00.1)2=811.8+0.01 {By using the identity(ab)2=a2+b22ab}=79.21(ix)10.5×9.5=10+0.5100.5=[102(0.5)2] {By using the identity(ab)(a+b)=a2b2}=100 0.25 = 99.75

Q.55 

Using a2 b2 = (a + b) (a b), find(i) 512 492 (ii) (1.02)2 (0.98)2 (iii) 1532 1472(iv) 12.12 7.92

Ans

(i)512492=(51+49)(5149)=100×2=200(ii) (1.02)2 (0.98)2 =(1.02+0.98)(1.020.98)=2×0.04=0.08(iii)15321472=(153+147)(153147)=300×6=1800(iv)12.127.92=(12.1+7.9)(12.17.9)=20.0×4.2=84

Q.56 

Using(x+a)(x+b)=x2+(a+b)x+ab,find(i)103×104(ii)5.1×5.2(iii)103×98(iv)9.7×9.8

Ans

(i)103×104=(100+3)(100+4)=(100)2+3+4×100+3×4=10000+700+12=10712(ii)5.1×5.2=(5+0.1)(5+0.2)=(5)2+0.1+ 0.2×5+0.1×0.2=25+1.5+0.02=26.52(iii)103×98=(100+3)(1002)=(100)2+[3+(2)]×100+3×(2)=10000+1006=10094(iv)9.7×9.8=(100.3)(100.2)=(10)2+[(0.3)+(0.2)]×10+(0.3)×(0.2)=100+(0.5)10+0.06=100.065=95.06

Q.57 For each of the given solid, the two views are given. Match for each solid the corresponding top and front views. The first one is done for you.

Ans

Q.58 For each of the given solid, the three views are given. Identify for each solid the corresponding top,

front and side views.

Ans

Q.59 For each given solid, identify the top view, front view and side view.

Ans

Q.60 Draw the front view, side view and top view of the given objects.

Ans

Q.61 Look at the given map of a city.

Answer the following.

  1. Colour the map as follows: Blue-water, red-fire station, orange-library, yellow – schools, Green – park, Pink – College, Purple – Hospital, Brown – Cemetery.
  2. Mark a green ‘X’ at the intersection of Road ‘C’ and Nehru Road, Green ‘Y’ at the intersection of Gandhi Road and Road A.
  3. In red, draw a short street route from Library to the bus depot.
  4. Which is further east, the city park or the market?
  5. Which is further south, the primary school or the Sr. Secondary School?

Ans

  1. Between the Market and the City Park, the City Park is further east.
  2. Between the Primary School and the Sr. Secondary School, the Sr. Secondary School is further south.

Q.62 Draw a map of your class room using proper scale and symbols for different objects.

Ans

Q.63 Draw a map of your school compound using proper scale and symbols for various features like play ground main building, garden etc.

Ans

Q.64 Draw a map giving instructions to your friend so that she reaches your house without any difficulty.

Ans

Q.65 Can a polyhedron have for its faces

(i) 3 triangles?
(ii) 4 triangles?
(iii) a square and four triangles?

Ans

(i) No, such a polyhedron is not possible. A polyhedron has minimum 4 faces.

(ii) Yes, a triangular pyramid has 4 triangular faces.

(iii) Yes, a square pyramid has a square face and 4 triangular faces.

Q.66 Is it possible to have a polyhedron with any given number of faces? (Hint: Think of a pyramid).

Ans

Yes, it is possible, if a polyhedron has minimum of 4 faces

Q.67 Which are prisms among the following?

Ans

(i) It is not a polyhedron as it has a curved surface. Therefore, it will not be a prism also.

(ii) It is a prism.

(iii)Since the faces are triangular, so it is a pyramid. Hence, it is not a prism

(iv) It is a prism.

Q.68 (i) How are prisms and cylinders alike?

(ii) How are pyramids and cones alike?

Ans

(i) A prism that has a circle as its base can be thought of as a circular prism.

(ii) A pyramid that has a circle as its base can be thought of as a circular pyramid.

Q.69 Is a square prism same as a cube? Explain.

Ans

A square prism has a square as its base. Its height is not necessarily same as the side of the square. Thus, it is not necessary that always a square prism is same as a cube. It can be cuboid also.

Q.70 Verify Euler’s formula for these solids.

Ans

The Euler’s formula F+V–E = 2

(i) Number of faces = F = 7

Number of vertices = V = 10

Number of edges = E = 15

We have, F + V − E = 7 + 10 − 15

= 17 − 15

= 2

Hence, verified.

(ii) Number of faces = F = 9

Number of vertices = V = 9

Number of edges = E = 16

F + V − E = 9 + 9 − 16

= 18 − 16

= 2

Hence, verified.

Q.71 Using Euler’s formula find the unknown.

Faces ? 5 20
Vertices 6 ? 12
Edges 12 9 ?

Ans

By Euler’s formula, we have F + V − E = 2

(i) F + 6 − 12 = 2

F − 6 = 2

F = 8

(ii) 5 + V − 9 = 2

V − 4 = 2

V = 6

(iii) 20 + 12 − E = 2

32 − E = 2

E = 30

Thus, the table can be completed as:

Faces 8 5 20
Vertices 6 6 12
Edges 12 9 30

Q.72 Can a polyhedron have 10 faces, 20 edges and 15 vertices?

Ans

Number of faces = F = 10

Number of edges = E = 20

Number of vertices = V = 15

By using Euler’s Formula, we can find whether a polyhedron satisfies the given conditions.

Therefore, we have, F + V − E = 2

But, F + V − E = 10 + 15 – 20

= 25 – 20

= 5 ≠ 2

Since Euler’s formula is not satisfied, such a polyhedron is not possible.

Q.73 A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area?

Ans

Perimeter of square=4×side = 4×60 m = 240 mPerimeter of rectangle=2(Length+Breadth) = 2(80 m+Breadth) = 160 m+2×Breadth Perimeter of the square=Perimeter of the rectangle 160 m+2×Breadth =240 mBreadth of the rectangle=(802)m=40 mArea of square=(Side)2=(60 m)2=3600 m2Area of rectangle=Length×Breadth =(80×40)m2 =3200 m2Therefore, the area of the square field is larger thanthe area of the rectangular field.

Q.74 Mrs. Kaushik has a square plot with the measurement as shown in the figure. She wants to construct a house in the middle of the plot. A garden is developed around the house. Find the total cost of developing a garden around the house at the rate of ₹ 55 per m2.

Ans

Area of the square plot=(25 )2m2=625 m2Area of the house=(15 m)×(20 m)=300 m2Area of the remaining portion=(Area of square plot)(Area of the house) =625 m2300 m2 = 325 m2Total cost of developing a garden around the house at the rate of Rs 55 per m2=(55×325) = 17,875

Q.75 The shape of a garden is rectangular in the middle and semi circular at the ends as shown in the diagram. Find the area and the perimeter of this garden [Length of rectangle is 20 – (3.5 + 3.5) metres].

Ans

Length of the rectangle=[20(3.5 + 3.5)]metres=13 mCircumference of 1 semicircular part=πr=227×3.5 =11 m.

Perimeter of the garden=AB+Length of both semicircular regions+CD = 13 m+22 m+13 m = 48 mArea of the garden=Area of rectangle+2×Area of two semicircular regions =[(13×7)+2×12×227×(3.5)2]m2 =(91+38.5)m2 =129.5m2Area and perimeter of the garden are 129.5m2 and 48 m.

Q.76 A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m2? (If required you can split the tiles in whatever way you want to fill up the corners).

Ans

Area of parallelogram=Base×HeightHence, area of one tile=24 cm×10 cm=240 cm2Required number of tiles=Areaofthe floorArea of eachtile=1080 m2240cm2 Since1m=100cm 1m2=10000cm21080 m2=10000×1080 cm2 =10,800,000cm2Hence,the required number of tiles =10,800,000  cm2240cm2 = 45000

Q.77 An ant is moving around a few food pieces of different shapes scattered on the floor. For which food-piece would the ant have to take a longer round? Remember, circumference of a circle can be obtained by using the expression c = 2πr, where r is the radius of the circle.

Ans

(a) Radius of semi-circular part={2.82}cm =1.4 cm Perimeter of the given figure=2.8 cm + πr =2.8 cm+(227×1.4)cm =2.8 cm+4.4cm =7.2cm (b) Radius of semi-circular part={2.82}cm=1.4 cm Perimeter of the given figure=1.5 cm+2.8 cm+1.5 cm+π(1.4 cm) =5.8 cm+227×1.4 cm =5.8 cm+4.4cm =10.2cm (c)Radius of semi-circular part={2.82}cm=1.4 cm Perimeter of the given figure=2 cm+πr+2 cm =4 cm+227×1.4 cm =4 cm+4.4cm =8.4cm Thus, the ant will have to take a longer round for the food piece ( b ), because the perimeter of the figure given in option ( b ) is the greatest among all. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@BD3F@

Q.78 The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are 1 m and 1.2 m and perpendicular distance between them is 0.8 m.

Ans

Area of trapezium=12(Sum of parallel sides)×(Distances between parallel sides) =[12(1+1.2)(0.8)]m2 =0.88m2

Q.79 The area of a trapezium is 34 cm2 and the length of one of the parallel sides is 10 cm and its height is 4 cm. Find the length of the other parallel side.

Ans

It is given that,area of trapezium=34 cm2and height=4 cmLet the length of one parallel side be a. We know that,Area of trapezium=12(Sum of parallel sides)×(Distances between parallel sides) 34 cm2=12(10cm+a)×4 cm34 cm2=2(10cm+a)17 cm=10cm+aa=17 cm10cm=7 cmThus, the length of the other parallel side is 7 cm.

Q.80 Length of the fence of a trapezium shaped field ABCD is 120 m. If BC = 48 m, CD = 17 m and AD = 40 m, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.

Ans

Length of the fence of a trapezium shaped field ABCD=AB+BC+CD+DA120 m=AB+48 m+17 m+40 mAB=120 m105 m=15 m Area of the field ABCD=12(AD+BC)×AB                            =12(40+48)×15m2                            =12×88×15m2                            =660m2Area of the field ABCD=660m2

Q.81 The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field.

Ans

It is given that,Length of the diagonal,d= 24 mLength of the perpendiculars,h1and h2, from the opposite verticesto the diagonal are 8 m and 13 mArea of the quadrilateral =12dh1+h2 =12×24×13 m+8 m =12×24×21 =252 m2Thus, the area of the field is 252 m2.

Q.82 The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.

Ans

Area of rhombus = 1 2 ( Product of its diagonals ) Therefore, area of the given rhombus= 1 2 ( 7.5×12 ) = 45 cm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@9114@

Q.83 Find the area of a rhombus whose side is 5 cm and whose altitude is 4.8 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal.

Ans

Let the other diagonal of the rhombus be x.Since rhombus is also a parallelogram.Therefore, area of the given rhombus=Base×Height =5 cm×4.8 cm =24 cm2Also, area of rhombus =12(Product of its diagonals)24 cm2=12(8 cm×x)x=24×28cm=6 cm The length of the other diagonal of the rhombus is 6 cm andarea of the given rhombus is 24 cm2.

Q.84 The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m2 is ₹ 4.

Ans

Area of rhombus = 12Product of its diagonalsArea of each tile = 12×45×30cm2 = 675 cm2Area of 3000 tiles = 675×3000cm2                        = 2025000 cm2                       = 202.5 m2the total cost of polishing the floor, if the cost per m2 is Rs 4= 4×202.5= 810Thus, the cost of polishing the floor is Rs 810.

Q.85 Mohan wants to buy a trapezium shaped field.

Its side along the river is parallel to and twice the side along the road. If the area of this field is 10500 m2 and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.

Ans

Let the length of the field along the road belm. Hence, the length of the field along the river will be 2lm. Area of trapezium = 1 2 ( Sum of parallel sides ) ( Distance between the parallel sides ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqabaWaaqaafaaakqaabeqaaiaabYeacaqGLbGaaeiDaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaaeiBaiaabwgacaqGUbGaae4zaiaabshacaqGObGaaeiiaiaab+gacaqGMbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGMbGaaeyAaiaabwgacaqGSbGaaeizaiaabccacaqGHbGaaeiBaiaab+gacaqGUbGaae4zaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaaeOCaiaab+gacaqGHbGaaeizaiaabccacaqGIbGaaeyzaiaaysW7caWGSbGaaGjbVlaab2gacaGGUaGaaeiiaaqaaiaabIeacaqGLbGaaeOBaiaabogacaqGLbGaaiilaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaaeiBaiaabwgacaqGUbGaae4zaiaabshacaqGObGaaeiiaiaab+gacaqGMbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGMbGaaeyAaiaabwgacaqGSbGaaeizaiaabccacaqGHbGaaeiBaiaab+gacaqGUbGaae4zaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaaeOCaiaabMgacaqG2bGaaeyzaiaabkhacaqGGaGaae4DaiaabMgacaqGSbGaaeiBaiaabccacaqGIbGaaeyzaiaabccacaqGYaGaamiBaiaaysW7caqGTbGaaiOlaaqaaiaabgeacaqGYbGaaeyzaiaabggacaqGGaGaae4BaiaabAgacaqGGaGaaeiDaiaabkhacaqGHbGaaeiCaiaabwgacaqG6bGaaeyAaiaabwhacaqGTbGaaeiiaiabg2da9maalaaabaGaaGymaaqaaiaaikdaaaWaaeWaaeaacaqGtbGaaeyDaiaab2gacaqGGaGaae4BaiaabAgacaqGGaGaaeiCaiaabggacaqGYbGaaeyyaiaabYgacaqGSbGaaeyzaiaabYgacaqGGaGaae4CaiaabMgacaqGKbGaaeyzaiaabohaaiaawIcacaGLPaaacaqGGaWaaeWaaqaabeqaaiaabseacaqGPbGaae4CaiaabshacaqGHbGaaeOBaiaabogacaqGLbGaaeiiaiaabkgacaqGLbGaaeiDaiaabEhacaqGLbGaaeyzaiaab6gaaeaacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabchacaqGHbGaaeOCaiaabggacaqGSbGaaeiBaiaabwgacaqGSbGaaeiiaiaabohacaqGPbGaaeizaiaabwgacaqGZbaaaiaawIcacaGLPaaaaaaa@E24E@ 10500 m2=12(l+2l) (100 m)3l=(2×10500100)m3l=210 ml =70 mThus, length of the field along the river=(2×70) m=140 m

Q.86 Top surface of a raised platform is in the shape of a regular octagon as shown in the figure. Find the area of the octagonal surface.

Ans


Side of regular octagon = 5 cm
Area of trapezium ABCH = Area of trapezium DEFG

Area of trapezium ABCH = 12×4×11+5m2                               = 12×4×16m2                               =  32 m2

Area of rectangle HGDC = 11 × 5 = 55 m2
Area of octagon = Area of trapezium ABCH + Area of trapezium DEFG + Area of rectangle HGDC
= 32 m2 + 32 m2 + 55 m2 = 119 m2
Therefore, the area of octagonal surface is 119 m2

Q.87 There is a pentagonal shaped park as shown in the figure.

For finding its area Jyoti and Kavita divided it in two different ways.

Find the area of this park using both ways. Can you suggest some other way of finding its area?

Ans

Jyoti way of calculating area:

Area of pentagon=2( Area of trapezium ABCF ) =[ 2× 1 2 ×(15+30)( 15 2 ) ] m 2 =337.5 m 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@78C6@

Kavita’s way of calculating area:

Area of pentagon ABCDE=Area of ΔABE+Area of square BCDE =[ 1 2 ×15×( 3015 )+ ( 15 ) 2 ] m 2 =[ 1 2 ×15×15+225 ] m 2 =337.5 m 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@9817@

No, there is no other way to find its area.

Q.88 Diagram of the adjacent picture frame has outer dimensions 24 cm × 28 cm and inner dimensions 16 cm × 20 cm. Find the area of each section of the frame, if the width of each section is same.

Ans

It is given that the width of each section is same.

Therefore,

LM = BM = CN = NE = OF = OH = PK = PI

CH = CN + NO+ OH

28 = 2CN+20

2CN = 28 – 20

2CN = 8

CN = 4 cm

Hence, LM = BM = NE = OF = OH = PK = PI = 4 cm

Area of section NDGO = Area of section AMPJ =[12(20+28)(4)]cm2 =96 cm2Area of section NDGO = Area of section AMPJ=96 cm2Also,Area of section AMND=Area of section PJGO =[12(16+24)(4)] =80 cm2Area of section AMND=Area of section PJGO=80 cm2

Q.89 There is a pentagonal shaped park as shown in the figure.

For finding its area Jyoti and Kavita divided it in two different ways.

Find the area of this park using both ways. Can you suggest some other way of finding its area?

Ans

Jyoti’s way of calculating area:

Area of pentagon=2(Area of trapezium ABCF) =[2×12×(15+30)(152)]m2 =337.5 m2

Kavita’s way of calculating area:

Area of pentagon ABCDE=Area of ΔABE+Area of square BCDE =[12×15×(3015)+(15)2]m2 =[12×15×15+225]m2 =337.5 m2

No, there is no other way to find its area.

Q.90 B

Ans

We know that,Total surface area of the cuboid=2 (lb+bh+hl)Total surface area of cuboid=[2{(60)(40)+(40)(50)+(50)(60)}]cm2 =[2(2400+2000+3000)] cm2 =[2(2400+2000+3000)](2×7400) cm2 =[2(2400+2000+3000)]14800 cm2Total surface area of the cube=6 (l)2 =6 (50 cm)2 =15000 cm2Since the total surface area of cuboid is less thanthe total surface area of cube,therefore, the cuboidal box willrequire lesser amount of material.

Q.90 A suitcase with measures 80 cm × 48 cm × 24 cm is to be covered with a tarpaulin cloth. How many metres of tarpaulin of width 96 cm is required to cover 100 such suitcases?

Ans

A suitcase is in the shape of a cuboid.Total surface area of a cuboid= 2(lb+bh+hl)Hence,Total surface area of suitcase=2[(80)(48)+(48)(24)+(24)(80)] cm2 =2[3840 + 1152 + 1920] cm2 =13824 cm2Total surface area of 100 suitcases=(13824 × 100)cm2 =1382400 cm2Required tarpaulin=Length×Breadth 1382400 cm2 =Length×96 cmLength=(138240096)cm=14400 cm=144 mThus, 144 m of tarpaulin is required to cover 100 suitcases.

Q.91 Find the side of a cube whose surface area is 600 cm2.

Ans

Given that, surface area of cube=600 cm2Let the length of each side of cube bex.Surface area of cube=6 (Side)2 600 cm2=6x2 x2=100 cm2 x=10 cm The side of the cube is 10 cm.

Q.92 Rukhsar painted the outside of the cabinet of measure 1 m × 2 m × 1.5 m. How much surface area did she cover if she painted all except the bottom of the cabinet?

Ans

Length of the cabinet = 2 mBreadth of the cabinet = 1 mHeight of the cabinet = 1.5 m Area of the cabinet that was painted=2h(l+b)+lb = [2×1.5×(2+1)+(2)(1)] m2 = [3(3)+2] m2 = (9+2) m2 = 11 m2 Area of the cabinet that was painted is 11 m2.

Q.93 Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth and height of 15 m, 10 m and 7 m respectively. From each can of paint 100 m2 of area is painted. How many cans of paint will she need to paint the room?

Ans

Given that,Length=15 m breadth=10 mheight=7 mArea of the hall to be painted=(Area of the walls)+(Area of the ceiling) = 2h(l+b)+lb=2(7)(15+10)+15×10 m2= 14(25)+150 m2= 500 m2It is given that 100 m2area can be painted from each can.Number of cans required to paint an area of 500 m2=500100=5Hence, 5 cans are required to paint the walls and the ceiling of the cuboidal hall.

Q.94 Describe how the two figures at the right are alike and how they are different. Which box has larger lateral surface area?

Ans

Both the figures have the same heights and the difference between the two figures is that one is a cylinderand the other is a cube. Now,Lateral surface area of the cube=4l2=4(7 cm)2=196 cm2Lateral surface area of the cylinder=2πrh sq. units =(2×227×72×7)cm2 =154 cm2Hence, the cube has larger lateral surface area.

Q.95 A closed cylindrical tank of radius 7 m and height 3 m is made from a sheet of metal. How much sheet of metal is required?

Ans

Total surface area of cylinder=2πr(r+h) sq. units =[2×227×7(7+3)]m2 =440 m2Thus, 440 m2 sheet of metal is required.

Q.96 The lateral surface area of a hollow cylinder is 4224 cm2. It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of rectangular sheet.

Ans

Surface area of hollow cylinder = Area of rectangular sheet4224 cm2=Length×33 cmLength =422433=128 cm The length of the rectangular sheet is 128 cm.Perimeter of the rectangular sheet=2(Length+Width) =[2(128+33)] cm =(2×161) cm =322 cmPerimeter of the rectangular sheet=322 cm.

Q.97 A road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is 84 cm and length is 1 m.

Ans

Given Diameter= 84 cmr = 42 cm=42100 mLength=1 m In one revolution, the roller will cover an area equal to itslateral surface area. Lateral surface area of cylinder=2πrh =2×227×42 cm×1 m =2×227×42 100m×1 m =264100 m2In 750 revolutions, area of the road covered=750×264100 m2 = 1980 m2

Q.98 A company packages its milk powder in cylindrical container whose base has a diameter of 14 cm and height 20 cm. Company places a label around the surface of the container (as shown in the figure). If the label is placed 2 cm from top and bottom, what is the area of the label?

Ans

Height of the label=20 cm2 cm2 cm=16 cmDiameter=14 cmRadius of the label=142 cm= 7 cmLabel is in the form of a cylinder having its radius and height as 7 cm and 16 cm. Area of the label=2πrh =(2×227×7×16)cm2 =704 cm2Area of the label=704 cm2

Q.99 Given a cylindrical tank, in which situation will you find surface area and in which situation volume.

(a) To find how much it can hold.

(b) Number of cement bags required to plaster it.

(c) To find the number of smaller tanks that can be filled with water from it.

Ans

(a) We will find the volume.

(b) We will find the surface area.

(c) We will find the volume.

Q.100 Diameter of cylinder A is 7 cm, and the height is 14 cm. Diameter of cylinder B is 14 cm and height is 7 cm. Without doing any calculations can you suggest whose volume is greater? Verify it by finding the volume of both the cylinders. Check whether the cylinder with greater volume also has greater surface area?

Ans

The heights and diameters of these cylinders A and B areinterchanged.If measures of ‘r’ and ‘h’ are same, then the cylinder withgreater radius will have greater volume.Radius of cylinder A =72cm Radius of cylinder B =142cm=7 cmAs the radius of cylinder B is greater, therefore, the volume of cylinder B will be greater.Let us verify it by calculating the volume of both the cylinders.Volume of cylinder A=πr2h=(227×72×72×14)cm3=539 cm3Volume of cylinder B=πr2h=(227×7×7×7)cm3=1078 cm3Hence,volume of cylinder B is greater. Surface area of cylinder A=2πr(h+r)=[227×72(72+14)]cm2=[22×(7+282)]cm2=385 cm2Surface area of cylinder B=2πr(h+r)=[227×7(7+7)]cm2=[44×14]cm2=616 cm2Thus, the surface area of cylinder B is also greater thanthe surface area of cylinder A.

Q.101 Find the height of a cuboid whose base area is 180 cm2 and volume is 900 cm3?

Ans

Base area of the cuboid=180 cm2 Length×Breadth=180 cm2 Volume of cuboid=Length×Breadth×Height 900 cm3=180 cm2×Height Height=900180=5 cmThus, the height of the cuboid is 5 cm.

Q.102 A cuboid is of dimensions 60 cm × 54 cm × 30 cm. How many small cubes with side 6 cm can be placed in the given cuboid?

Ans

Volume of cuboid=60 cm×54 cm×30 cm=97200 cm3Side of the cube=6 cmVolume of the cube=(6)3cm3=216 cm3Required number of cubes =Volume of the cuboidVolume of the cube =97200 cm3216 cm3=450450 cubes can be placed in the given cuboid.

Q.103 Find the height of the cylinder whose volume is 1.54 m3 and diameter of the base is 140 cm?

Ans

Given,Volume of cylinder=1.54 m3Diameter of the base = 140 cmRadius of the base= (1402) cm=70 cm = 70100mVolume of cylinder=πr2h1.54 m3=227×70100×70100×hh=(1.54×10022×7)m=1 mThus, the height of the cylinder is 1 m.

Q.104 A milk tank is in the form of cylinder whose radius is 1.5 m and length is 7 m. Find the quantity of milk in litres that can be stored in the tank?

Ans

Radius of cylinder = 1.5 mLength of cylinder = 7 mVolume of cylinder=πr2h =227×1.5×1.5×7 =49.5 m3We know that ,1m3= 1000 LThe quantity of milk in liters=(49.5 × 1000)L = 49500 LTherefore, 49500 L of milk can be stored in the tank.

Q.105 If each edge of a cube is doubled,

(i) how many times will its surface area increase?

(ii) how many times will its volume increase?

Ans

Let initially the edge of the cube be ‘x’.(i) Initial surface area=6x2If each edge of the cube is doubled, then it becomes 2x. New surface area=6(2x)2 =24x2 =4×6x2Hence, the surface area will be increased by 4 times.(ii) Initial volume of the cube =x3When each edge of the cube is doubled, it becomes 2x.New volume=(2x)3=8x3=8×x3 The volume of the cube will be increased by 8 times.

Q.106 Water is pouring into a cuboidal reservoir at the rate of 60 litres per minute. If the volume of reservoir is 108 m3, find the number of hours it will take to fill the reservoir.

Ans

Volume of cuboidal reservoir=108 m3 =(108×1000) L =108000 LIt is given that water is being poured at the rate of 60 L per min.i.e.(60×60) L=3600 L per hourRequired number of hours=1080003600= 30 hoursIt will take 30 hours to fill the reservoir.

Q.107 

Evaluate.(i) 3-2 (ii) (-4) -2 (iii) (12)-5

Ans

i 32=132=19[By using am=1am]ii (-4) -2=1(4)2=116 [By using am=1am]iii (12)5=125=25=2×2×2×2×2=32

Q.108 

Simplify and express the result in power notation with positive exponent.(i) (- 4)5 ÷ (- 4)8 (ii) (123)2 (iii) (– 3)4×(53)4 (iv) (3– 7÷ 3 – 10) × 3 – 5 (v) 2 – 3 × (- 7)– 3

Ans

(i) (- 4)5 ÷ (- 4)8=(- 4)5-8[By using am÷an=am-n]=(- 4)-3=1(-4)3[By using a-m=1am](ii)(123)2=1(23)2=126 [By using (am)n=amn] (iii) (3)4×(53)4=(1)4×34×5434=(1)4×54=54 [By using (1)4=1] (iv) (3– 7÷ 3 – 10) × 3 – 5=3(– 7)(-10) × 3 – 5 [By using am÷an=am-n]=33 × 3 – 5=3(3)+(-5) [By using am×an=am+n]=3-2=132 [By using a-m=1am](v) 2 – 3 × (- 7)-3=123×1(-7)3=1[(-7)]3 [By using am×bm=abm]=1(-14)3

Q.109 

Find the value of:(i) (30 + 41) × 22(ii)(21 × 41) ÷ 22iii(12)2+(13)2+(14)2(iv) (3-1 + 4-1 + 5-1)0 v {(23)2}2

Ans

(i)(3°+41)×22=(1+14)×22 [By using a0=1, am=1am]=54×4=5(ii)(21×41)÷22=(21×(22)1)÷22 =21×(22)÷22 [By using (am)n=amn]=21+(2)÷22 [By using am×an=am+n] =2-3÷2-2=2-3-(-2) [By using am÷an=am-n]=2-3+2=2-1=12(iii) (12)-2+(13)-2+(14)-2=(21)2+(31)2+(41)2  [By using a-m=1am]=22+32+42=4+9+16=29(iv) (3-1 + 4-1 + 5-1)0 =(13+14+15)0 [By using a-m=1am]=1 [By using a0=1] v {(23)2}2={(32)2}2 [By using am=1am]={32(2)2}2 [By using (ab)m=ambm]=(94)2=8116

Q.110 

Evaluatei81×5324 ii(51×21)×61

Ans

i 81×5324=24×5381[By using am=1am]=24×5323 =243×53 [By using am÷an=amn] =2×125=250 (ii)(51×21)×61=(15×12)×16[By using am=1am]=110×16=160

Q.111 

Find the value of m for which 5m ÷ 53 = 55.

Ans

5m ÷ 5-3 = 555m-(-3)= 55[By using am÷an=am-n]5m+3= 55m+3=5 (Since the bases are same,  their powers must be equal.)m=53m=2

Q.112 Evaluate

i {(13)1(14)1}1ii (58)7× (85)4

Ans

i {(13)-1(14)-1}-1= {(31)(41)}-1 [By using a-m=1am]= {3-4}-1= (-1)-1= 1(-1)= -1 ii (58)7×(85)4=5787×8454 [By using (ab)m=ambm]

=8757×5484[By using am=1am]=874574 [By using am÷an=amn]=8353=512125

Q.113 Simplify

(i) 25×t-45-3×10×t-8 (t ≠ 0) (ii) 3-5×10-5×1255-7×6-5

Ans

i25×t453×10×t8 t0=52×t453×5×2×t8=52×t453+1×2×t8 [By using : am×an=am+n]=52×t452×2×t8 =52+2×t4+8 2 [By using : am÷an=amn]=54×t4 2=625t42 ii35×105×12557×65=35×(2×5)5×5357×(2×3)5=35+5×25+5×55+3+7[By using : am÷an=amn]=30×20×55[By using : a0=a1]=55

Q.114 Express the following numbers in standard form.

(i) 0.0000000000085 (ii) 0.00000000000942

(iii) 6020000000000000 (iv) 0.00000000837

(v) 31860000000

Ans

The standard form of a number is

N ×10power, where N is number greater than orequal to 1 but less than 10. (i) The standard form of 0.0000000000085 = 8.5 × 1012(ii) The standard form of 0.00000000000942 = 9.42 × 1012(iii) The standard form of 6020000000000000= 6.02 × 1015(iv) The standard form of 0.00000000837 = 8.37 × 109(v) The standard form of 31860000000 = 3.186 × 1010

Q.115 Express the following numbers in usual form.

(i) 3.02 × 10– 6 (ii) 4.5 × 104 (iii) 3 × 10– 8

(iv) 1.0001 × 109 (v) 5.8 × 1012 (vi) 3.61492 × 106

Ans

(i) 3.02 × 106=3.02106=302×1021000000= 0.00000302(ii) 4.5 × 104= 4.5×10000=45000(iii) 3 × 108= 3108=0.00000003 (iv) 1.0001 × 109= 1000100000(v) 5.8 × 1012= 5800000000000(vi) 3.61492 × 106= 3614920

Q.116 Express the number appearing in the following statements in standard form.

(i) 1 micron is equal to 1/1000000 m.

(ii) Charge of an electron is 0.000,000,000,000,000,000,16 coulomb.

(iii) Size of a bacteria is 0.0000005 m

(iv) Size of a plant cell is 0.00001275 m

(v) Thickness of a thick paper is 0.07 mm

Ans

The standard form the numbers are as follows:

(i)11000000=1106= 1 × 106(ii) 0.000,000,000,000,000,000,16=1.6 × 101020=1.6 × 1019(iii) 0.0000005 =510000000=5107= 5 × 107(iv) 0.00001275 =1.275×103108= 1.275 × 105(v) 0.07 =7100= 7 × 102

Q.117 In a stack there are 5 books each of thickness 20mm and 5 paper sheets each of thickness 0.016 mm. What is the total thickness of the stack?

Ans

Thickness of each book=20 mm Hence, thickness of 5 books=( 5×20 ) mm=100 mm Thickness of each paper sheet=0.016 mm Hence, thickness of 5 paper sheets=( 5 × 0.016 ) mm=0.080 mm Total thickness of the stack=Thickness of 5 books+Thickness of 5 paper sheets = ( 100+0.080 )mm=100.08 mm=1.0008×1 0 2 mm MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiGacaGaaeqabaWaaqaafaaakqaabeqaaiaabsfacaqGObGaaeyAaiaabogacaqGRbGaaeOBaiaabwgacaqGZbGaae4CaiaabccacaqGVbGaaeOzaiaabccacaqGLbGaaeyyaiaabogacaqGObGaaeiiaiaabkgacaqGVbGaae4BaiaabUgacqGH9aqpcaqGYaGaaGimaiaabccacaqGTbGaaeyBaaqaaiaabIeacaqGLbGaaeOBaiaabogacaqGLbGaaiilaiaabccacaqG0bGaaeiAaiaabMgacaqGJbGaae4Aaiaab6gacaqGLbGaae4CaiaabohacaqGGaGaae4BaiaabAgacaqGGaGaaeynaiaabccacaqGIbGaae4Baiaab+gacaqGRbGaae4Caiabg2da9maabmaabaGaaeynaiabgEna0kaabkdacaaIWaaacaGLOaGaayzkaaGaaeiiaiaab2gacaqGTbGaeyypa0JaaeymaiaaicdacaaIWaGaaeiiaiaab2gacaqGTbaabaGaaeivaiaabIgacaqGPbGaae4yaiaabUgacaqGUbGaaeyzaiaabohacaqGZbGaaeiiaiaab+gacaqGMbGaaeiiaiaabwgacaqGHbGaae4yaiaabIgacaqGGaGaaeiCaiaabggacaqGWbGaaeyzaiaabkhacaqGGaGaae4CaiaabIgacaqGLbGaaeyzaiaabshacqGH9aqpcaaIWaGaaiOlaiaaicdacaqGXaGaaeOnaiaabccacaqGTbGaaeyBaaqaaiaabIeacaqGLbGaaeOBaiaabogacaqGLbGaaiilaiaabccacaqG0bGaaeiAaiaabMgacaqGJbGaae4Aaiaab6gacaqGLbGaae4CaiaabohacaqGGaGaae4BaiaabAgacaqGGaGaaeynaiaabccacaqGWbGaaeyyaiaabchacaqGLbGaaeOCaiaabccacaqGZbGaaeiAaiaabwgacaqGLbGaaeiDaiaabohacqGH9aqpdaqadaqaaiaabwdacaqGGaGaey41aqRaaeiiaiaaicdacaGGUaGaaGimaiaabgdacaqG2aaacaGLOaGaayzkaaGaaeiiaiaab2gacaqGTbGaeyypa0JaaGimaiaac6cacaaIWaGaaeioaiaaicdacaqGGaGaaeyBaiaab2gaaeaacaqGubGaae4BaiaabshacaqGHbGaaeiBaiaabccacaqG0bGaaeiAaiaabMgacaqGJbGaae4Aaiaab6gacaqGLbGaae4CaiaabohacaqGGaGaae4BaiaabAgacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabohacaqG0bGaaeyyaiaabogacaqGRbGaeyypa0JaaeivaiaabIgacaqGPbGaae4yaiaabUgacaqGUbGaaeyzaiaabohacaqGZbGaaeiiaiaab+gacaqGMbGaaeiiaiaabwdacaqGGaGaaeOyaiaab+gacaqGVbGaae4AaiaabohacqGHRaWkcaqGubGaaeiAaiaabMgacaqGJbGaae4Aaiaab6gacaqGLbGaae4CaiaabohacaqGGaGaae4BaiaabAgacaqGGaaabaGaaeynaiaabccacaqGWbGaaeyyaiaabchacaqGLbGaaeOCaiaabccacaqGZbGaaeiAaiaabwgacaqGLbGaaeiDaiaabohaaeaacqGH9aqpcaqGGaWaaeWaaeaacaqGXaGaaGimaiaaicdacqGHRaWkcaaIWaGaaiOlaiaaicdacaqG4aGaaGimaaGaayjkaiaawMcaaiaab2gacaqGTbGaeyypa0JaaeymaiaaicdacaaIWaGaaiOlaiaaicdacaqG4aGaaeiiaiaab2gacaqGTbGaeyypa0Jaaeymaiaac6cacaaIWaGaaGimaiaaicdacaqG4aGaey41aqRaaeymaiaaicdadaahaaWcbeqaaiaabkdaaaGccaqGTbGaaeyBaaaaaa@2CF3@

Q.118 1. The following graph shows the temperature of a patient in a hospital, recorded every hour.
(a) What was the patient’s temperature at 1 p.m.?
(b) When was the patient’s temperature 38.5° C?
(c) The patient’s temperature was the same two times during the period given. What were these two times?
(d) What was the temperature at 1.30 p.m.? How did you arrive at your answer?
(e) During which periods did the patients’ temperature showed an upward trend?

Ans

(a) The patient’s temperature was 36.5°C at 1 p.m.
(b) At 12 noon, the patient’s temperature was 38.5°C.
(c) The two times when the patient’s temperature was same, were 1 p.m. and 2 p.m.
(d) The temperature at 1:30 p.m. is 36.5°C.
(e) The patient’s temperature showed an upward trend during the following periods: 9 a.m. to 10 a.m., 10 a.m. to 11 a.m., 2 p.m. to 3 p.m.

Q.119 The following line graph shows the yearly sales figures for a manufacturing company.
(a) What were the sales in (i) 2002 (ii) 2006?
(b) What were the sales in (i) 2003 (ii) 2005?
(c) Compute the difference between the sales in 2002 and 2006.
(d) In which year was there the greatest difference between the sales as compared to its previous year?

Ans

(a)
(i) The sales in 2002 were ₹ 4 crores.
(ii) The sales in 2006 were ₹ 8 crores.
(b)
(i) The sales in 2003 were ₹ 7 crores.
(ii) The sales in 2005 were ₹ 10 crores.
(c)
(i) The difference between the sales in 2002 and 2006 = ₹ (8 − 4) crores = ₹ 4 crores
(d) For finding the greatest difference between the sales as compared to its previous year, we have to find compare all the years.
Therefore, the difference between the sales of the year 2006 and 2005 = ₹ (10 − 8) crores = ₹ 2 crores
Difference between the sales of the year 2005 and 2004 = ₹ (10 − 6) crores = ₹ 4 crores
Difference between the sales of the year 2004 and 2003 = ₹ (7 − 6) crore = ₹ 1 crore
Difference between the sales of the year 2003 and 2002 = ₹ (7 − 4) crores = ₹ 3 crores
Hence, the difference was the maximum in the year 2005 as compared to its previous year 2004.

Q.120 For an experiment in Botany, two different plants, plant A and plant B were grown under similar laboratory conditions. Their heights were measured at the end of each week for 3 weeks. The results are shown by the following graph.
(a) How high was Plant A after (i) 2 weeks (ii) 3 weeks?
(b) How high was Plant B after (i) 2 weeks (ii) 3 weeks?
(c) How much did Plant A grow during the 3rd week?
(d) How much did Plant B grow from the end of the 2nd week to the end of the 3rd week?
(e) During which week did Plant A grow most?
(f) During which week did Plant B grow least?
(g) Were the two plants of the same height during any week shown here? Specify.

Ans

(a)
(i) The height of plant A was 7 cm after 2 weeks.
(ii) The height of plant A was 9 cm after 3 weeks.
(b)
(i) The height of plant B was 7 cm after 2 weeks.
(ii) The height of plant B was 10 cm after 3 weeks.
(c) The growth of plant A during 3rd week = 9 cm − 7 cm = 2 cm
(d) The growth of plant B from the end of the 2nd week to the end of the 3rd week = 10 cm − 7 cm = 3 cm
(e) The growth of plant A during 1st week = 2 cm − 0 cm = 2 cm
The growth of plant A during 2nd week = 7 cm − 2 cm = 5 cm
The growth of plant A during 3rd week = 9 cm − 7 cm = 2 cm
Therefore, plant A grew the most, i.e. 5 cm, during the 2nd week.

(f) The growth of the plant B during 1st week = 1 cm − 0 cm = 1 cm
The growth of the plant B during 2nd week = 7 cm − 1 cm = 6 cm
The growth of plant B during 3rd week = 10 cm − 7 cm = 3 cm
Therefore, plant B grew the least, during the 1st week.

(g) The two plants were of the same height during the 2nd week.

Q.121 The following graph shows the temperature forecast and the actual temperature for each day of a week.
(a) On which days was the forecast temperature the same as the actual temperature?
(b) What was the maximum forecast temperature during the week?
(c) What was the minimum actual temperature during the week?
(d) On which day did the actual temperature differ the most from the forecast temperature?

Ans

(a) The forecast temperature was same as the actual temperature on Tuesday, Friday, and Sunday.
(b) The maximum forecast temperature during the week was 35°C.
(c) The minimum actual temperature during the week was 15°C.
(d) The actual temperature differs the most from the forecast temperature on Thursday.

Q.122 Use the tables below to draw linear graphs.

(a) The number of days a hill side city received snow in different years.

Year 2003 2004 2005 2006
Days 8 10 5 12

(b) Population (in thousands) of men and women in a village in different years.

Year 2003 2004 2005 2006 2007
Number of Men 12 12.5 13 13.2 13.5
Number of Women 11.3 11.9 13 13.6 12.8

Ans

  1. Year 2003 2004 2005 2006
    Days 8 10 5 12

  2. Year 2003 2004 2005 2006 2007
    Number of Men 12 12.5 13 13.2 13.5
    Number of Women 11.3 11.9 13 13.6 12.8

Q.123 A courier-person cycles from a town to a neighbouring suburban area to deliver a parcel to a merchant. His distance from the town at different times is shown by the following graph.
(a) What is the scale taken for the time axis?
(b) How much time did the person take for the travel?
(c) How far is the place of the merchant from the town?
(d) Did the person stop on his way? Explain.
(e) During which period did he ride fastest?

Ans

(a) Scale taken for the time axis is 4 units = 1 hour
(b)

The person took 312hours to travel.

(c) The merchant is 22 km far from the town.
(d) Yes, the person stopped on his way from 10 a.m. to 10: 30 a.m. It is clearly shown by the horizontal line in the graph.
(e) From 8 a.m. to 9 a.m., the person travelled the maximum distance. So, the person’s ride was the fastest between 8 a.m. and 9 a.m.

Q.124 Can there be a time-temperature graph as follows? Justify your answer.

Ans

(i) As the temperature can increase with the increase in time, so it can be a time−temperature graph..
(ii) As the temperature can decrease with the decrease in time, so it can be a time−temperature graph.
(iii) Since different temperatures at the same time are not possible, so it cannot be a time−temperature graph.
(iv) As same temperature at different times is possible, so it can be a time−temperature graph.

Q.125 Plot the following points on a graph sheet. Verify if they lie on a line
(a) A(4, 0), B(4, 2), C(4, 6), D(4, 2.5)
(b) P(1, 1), Q(2, 2), R(3, 3), S(4, 4)
(c) K(2, 3), L(5, 3), M(5, 5), N(2, 5)

Ans

When we plot the points on the graph and then join them, it can be observed that they lie on the same line.

It can be observed from the graph that the points P, Q, R and S lie on the same line.

When all the points are joined, it can be observed that they don’t lie on the same line.

Q.126 Draw the line passing through (2, 3) and (3, 2). Find the coordinates of the points at which this line meets the x-axis and y-axis.

Ans

The line passing through (2, 3) and (3,2) meets the x-axis and y-axis are (5, 0) and (0, 5).

Q.127 Write the coordinates of the vertices of each of these adjoining figures.

Ans

The coordinates of the vertices in the given figure are as follows.

OABC: O (0, 0), A (2, 0), B (2, 3), C (0, 3)

PQRS: P (4, 3), Q (6, 1), R (6, 5), S (4, 7)

MLK : K (10, 5), L (7, 7), M (10, 8)

Q.128 State whether True or False. Correct that are false.
(i) A point whose x coordinate is zero and y-coordinate is non-zero will lie on the y-axis.
(ii) A point whose y coordinate is zero and x-coordinate is 5 will lie on y-axis.
(iii) The coordinates of the origin are (0, 0).

Ans

(i) True
(ii) False: The point whose y-coordinate is zero and x-coordinate is 5 will lie on x-axis.
(iii) True

Q.129 1. Draw the graphs for the following tables of values, with suitable scales on the axes.

(a) Cost of apples

Number of apples 1 2 3 4 5
Cost 5 10 15 20 25

(b) Distance travelled by a car

Time (in hours) 6a.m. 7 a.m. 8 a.m. 9 a.m.
Distances (in km) 40 80 120 160
  1. How much distance did the car cover during the period 7.30 a.m. to 8 a.m.?
  2. What was the time when the car had covered a distance of 100 km since it’s start?

(c) Interest on deposits for a year.

Number of apples 1 2 3 4 5
Cost 5 10 15 20 25
  1. Does the graph pass through the origin?
  2. Use the graph to find the interest on ₹ 2500 for a year.
  3. To get an interest of ₹ 280 per year, how much money should be deposited?

Ans

(a)

Number of apples 1 2 3 4 5
Cost 5 10 15 20 25

Scale:
X-axis, 1 unit = 10 apples
Y-axis, 1 unit = Rs 10
A graph of the given data is as follows.

(b)

Time (in hours) 6a.m. 7 a.m. 8 a.m. 9 a.m.
Distances (in km) 40 80 120 160

Scale:
X-axis, 2 units = 1 hour
Y-axis, 2 units = 40 km
A graph of the given data is as follows.

  1. The car covered a distance of 20 km during the period 7:30 a.m. to 8 a.m.
  2. The car covered a distance of 100 km since its start at 7:30 a.m.

(c)

Deposit (in Rs) 1000 2000 3000 4000 5000
Simple Interest (in Rs) 80 160 240 320 400

  1. Yes, the graph passes through the origin.
  2. The interest on Rs 2500 for a year is Rs 200.
  3. To get an interest of Rs 280 per year, Rs 3500 should be deposited.

Q.130 Draw a graph for the following.

(i)

Side of square 2 3 3.5 5 6
Perimeter 8 12 14 20 24

Is it a linear graph?

(ii)

Side of square (in cm) 2 3 3.5 5 6
Area (in cm2) 4 9 16 25 36

Is it a linear graph?

Ans

(i)

Side of square 2 3 3.5 5 6
Perimeter 8 12 14 20 24

Scale:
X axis: 1 unit = 5 cm
Y axis: 1 unit = 5 cm
It is a linear graph.

(ii)

Side of square (in cm) 2 3 3.5 5 6
Area (in cm2) 4 9 16 25 36

Scale:
X axis: 1 unit = 1 cm
Y axis: 1 unit = 5 cm2
It is not a linear graph.

Q.131

A mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of base. In the following table, find the parts of base that need to be added.Parts of red pigment1471220Parts of base8

Ans

The given mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of base. Since the ratio of red pigment and the base is same every time, therefore, the parts of red pigments and the parts of base are in direct proportion.

Let us take the unknown values as x1, x2, x3 and x4.
The given information in the form of a table is as follows.

Parts of red pigment1471220Parts of base8x1x2x3x4 According to direct proportion,18=4x1x1=8×4x1=3218=7x2x2=8×7x2=5618=12x3x3=8×12x3=9618=20x4x4=8×20x4=160

Therefore, the parts of a base to be added are shown as follows:

Parts of red pigment1471220Parts of base8325696160

Q.132 In Question 2 above, if 1 part of a red pigment requires 75 mL of base, how much red pigment should we mix with 1800 mL of base?

Ans

Let the parts of red pigment required to mix with 1800 mL of base be x.

The given information in the form of a table is as follows.

Parts of red pigment1xParts of base(in mL)751800

The parts of red pigment and the parts of base are in direct proportion.
Therefore, we obtain

According to direct proportion,175=x180075x=1800x=24

Thus, 24 parts of red pigments should be mixed with 1800 mL of base.

Q.133 A machine in a soft drink factory fills 840 bottles in six hours. How many bottles will it fill in five hours?

Ans

Let the number of bottles filled by the machine in five hours be x.

The given information in the form of a table is as follows.

Number of bottles840xTime taken(in hours)65

The number of bottles and the time taken to fill these bottles are in direct proportion. Therefore, we obtain

8406=x56x=840×56x=4200x=700

Thus, 700 bottles will be filled in 5 hours.

Q.134 A photograph of a bacterium enlarged 50,000 times attains a length of 5 cm as shown in the diagram. What is the actual length of the bacterium? If the photograph is enlarged 20,000 times only, what would be its enlarged length?

Ans

Let the actual length of the bacterium be x cm.
The given information in the form of a table is as follows.

Length of the bacterium (in cm) 5 x Number of times photograph of the bacterium was enlarged 50000 1 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@8C3C@

The number of times the photograph of the bacterium was enlarged and the lengths of the bacterium are in direct proportion.
Therefore, we obtain

5 50000 = x 1 50000x=5 x= 1 10000 x= 10 4 Hence, the actual length of the bacterium is 1 0 4 cm. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@867C@

Now, we have to find its enlarged length if the photograph is enlarged 20,000 times only.

Let the enlarged length of the bacterium be y cm.
The given information in the form of a table is as follows.

Length of the bacterium (in cm)5yNumber of times photograph of the bacterium was enlarged5000020000

The number of times the photograph of the bacterium was enlarged and the lengths of bacterium are in direct proportion.

Therefore, we obtain

550000=y20000100000=50000yy=10000050000y=2Hence, the enlarged length of the bacterium is 2 cm.

Q.135 In a model of a ship, the mast is 9 cm high, while the mast of the actual ship is 12 m high. If the length of the ship is 28 m, how long is the model ship?

Ans

Let the length of the mast of the model ship be x cm.
The given information in the form of a table is as follows:

Height of mastLength of shipModel of ship9xActual ship1228

The dimensions of the actual ship and the model ship are directly proportional to each other.

Therefore, we obtain:

912=x289×28=12xx=25212x=21

Thus, the length of the model ship is 21 cm.

Q.136 Suppose 2 kg of sugar contains 9 × 106 crystals. How many sugar crystals are there in
(i) 5 kg of sugar? (ii) 1.2 kg of sugar?

Ans

(i) Let the number of sugar crystals in 5 kg of sugar be x.
The given information in the form of a table is as follows:

Amount of sugar(in kg)25Number of crystals9×106x

The amount of sugar and the number of crystals it contains are directly proportional to each other.
Therefore, we obtain

29×106=5x2x=5×9×106x=5×9×1062x=22.5×106x=2.25×107

Hence, the number of sugar crystals is 2.25 × 107.

(ii) Let the number of sugar crystals in 1.2 kg of sugar be y.

The given information in the form of a table is as follows:

Amount of sugar(in kg)21.2Number of crystals9×106y

The amount of sugar and the number of crystals it contains are directly proportional to each other

.

29×106=1.2y2y=1.2×9×106x=1.2×9×1062x=5.4×106Hence, the number of sugar crystals is 5.4×106.

Therefore, we obtain

Q.137 Rashmi has a road map with a scale of 1 cm representing 18 km. She drives on a road for 72 km. What would be her distance covered in the map?

Ans

Let the distance represented on the map be x cm.
The given information in the form of a table is as follows.

Distance covered on map(in cm)1xDistance covered on road(in km)1872

The distances covered on road and represented on map are directly proportional to each other.
Therefore, we obtain

118=x7272=18xx=7218x=4

Hence, the distance represented on the map is 4 cm.

Q.138 A 5 m 60 cm high vertical pole casts a shadow 3 m 20 cm long. Find at the same time
(i) the length of the shadow cast by another pole 10 m 50 cm high
(ii) the height of a pole which casts a shadow 5m long.

Ans

Let the length of the shadow of the other pole be x m.
The given information in the form of a table is as follows.

Height of the pole(in m)5.6010.50Length of the shadow(in m)3.20x

The height of an object and length of its shadow are directly proportional to each other.
Therefore, we obtain

5.603.20=10.50x5.60x=10.50×3.20x=33.65.60x=6

Hence, the length of the shadow will be 6 m.

(ii) Let the height of the pole be y m.
The given information in the form of a table is as follows.

Height of the pole(in m)5.60yLength of the shadow(in m)3.205

The height of an object and length of its shadow are directly proportional to each other.
Therefore, we obtain

5.603.20=y55.60×5=3.20yy=5.60×53.20y=8.75

Thus, the height of the pole is 8.75 m or 8 m 75 cm.

Q.139 A loaded truck travels 14 km in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?

Ans

Let the distance travelled by the truck in 5 hours be x km.
We know, 1 hour = 60 minutes
∴ 5 hours = (5 × 60) minutes = 300 minutes
The given information in the form of a table is as follows.

Distance travelled(in km)14xTime(in min)25300

The distance travelled by the truck and the time taken by the truck are directly proportional to each other.

Therefore, we obtain

1425=x300300×14=25xx=420025x=168

Hence, the distance travelled by the truck is 168 km.

Q.140 Which of the following are in inverse proportion?

(i) The number of workers on a job and the time to complete the job.

(ii) The time taken for a journey and the distance travelled in a uniform speed.

(iii) Area of cultivated land and the crop harvested.

(iv) The time taken for a fixed journey and the speed of the vehicle.

(v) The population of a country and the area of land per person.

Ans

(i) It is in inverse proportion because if there are more workers, then it will take less time to complete that job.

(ii) No, these are not in inverse proportion because in more time, we may cover more distance with a uniform speed.

(iii) No, these are not in inverse proportion because in more area, more quantity of crop may be harvested.

(iv) It is in inverse proportion because with more speed, we may complete a certain distance in a lesser time.

(v) It is in inverse proportion because if the population is increasing/decreasing, then the area of the land per person will be decreasing/increasing accordingly.

Q.141 In a Television game show, the prize money of ₹ 1, 00,000 is to be divided equally amongst the winners. Complete the following table and find whether the prize money given to an individual winner is directly or inversely proportional to the number of winners?

Number of winners 1 2 4 5 8 10 20
Prize of each winner (in ₹ ) 1,00,000 50,000

Ans

Let the unknown prizes be x1, x2, x3, x4 and x5.

The given information is represented in a table below:

Number of winners 1 2 4 5 8 10 20
Prize of each winner (in ₹ ) 1,00,000 50,000 x1 x2 x3 x4 x5

From the table, we obtain

1 × 1,00,000 = 2 × 50,000 = 1,00,000

Thus, the number of winners and the amount given to each winner are inversely proportional to each other.

Therefore, we obtain

1×1,00,000=4×x1x1=1,00,0004=25,0001×1,00,000=5×x2x2=1,00,0005=