NCERT Solutions for Class 8 Maths Ch 14 Factorization Exercise 14.1

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NCERT Solutions for Class 8 Maths Ch 14 Factorization Exercise 14.1

NCERT Solutions for Class 8 Maths Chapter 14 Factorization Exercise 14.1 is provided here which are useful in fetching good marks in the examinations. If you get stuck in a questions then you can take help from the NCERT Solutions for Class 8 Maths. You can also download Studyrankers app from Playstore to view NCERT Solutions offline at your ease.

In exercise 14.1, you need to find the common factors of the given terms.

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NCERT Solutions for Class 8 Maths Ch 14 Factorization Exercise 14.2

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NCERT Solutions for Class 8 Maths Ch 14 Factorization Exercise 14.2

Chapter 14 Factorization Exercise 14.2 Class 8 Maths NCERT Solutions is very helpful in preparing yourself well in the examinations. These NCERT Solutions are prepared by Studyrankers experts who have in depth knowledge of the subject. Through the NCERT Solutions for Class 8 Maths, you can find the solutions of difficult problems and clear your doubts.

You can also download Studyrankers app from Playstore to view study materials for Class 8 offline. In Exercise 14.2, you have to factorise the given expressions.

NCERT Solutions for Class 8 Maths Ch 14 Factorization Exercise 14.3

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NCERT Solutions for Class 8 Maths Ch 14 Factorization Exercise 14.3

Chapter 14 Factorization Exercise 14.3 NCERT Solutions for Class 8 Maths is useful in the completing homework in no time and passing exam with flying colours. These NCERT Solutions are useful in clearing doubts whenever you find difficulty in solving any question. You can also download Studyrankers app from Playstore to view study materials for Class 8 Maths without any internet connection.

In Exercise 14.3, you have to divide the given polynomial by the given monomial.

NCERT Solutions for Class 8 Maths| Updated 2020-21

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NCERT Solutions for Class 8 Maths - Free PDF Download

NCERT Solutions for Class 8 Maths is one of the best way through which you can have glance of important topics. Finding accurate CBSE Class 8 Maths NCERT Solutions can be really tough. Thus, we have created detailed explanation of every question of NCERT Solutions for Class 8 Mathematics that will be helpful in completing your homework and learning basic concepts. It will help every Class 8 students so you can practice the questions and get your doubts cleared.

Studyrankers has prepared chapterwise NCERT Solutions for Class 8 Maths that are detailed so you can match you own answers which will surely helpful in improving marks in the examinations. Whenever you get stuck you can NCERT questions and answers easily which are updated as per the latest syllabus 2020-21 released by CBSE. You can find NCERT Solutions chapterwise which you can find by clicking on the chapter name.

Chapterwise NCERT Solutions for Class 8 Maths

• NCERT Solutions for Chapter 1 - Rational Numbers
• NCERT Solutions for Chapter 2 - Linear Equations in One Variable
• NCERT Solutions for Chapter 3 - Understanding Quadrilaterals
• NCERT Solutions for Chapter 4 - Practical Geometry
• NCERT Solutions for Chapter 5 - Data Handling
• NCERT Solutions for Chapter 6 - Square and Square Roots
• NCERT Solutions for Chapter 7 - Cube and Cube Roots
• NCERT Solutions for Chapter 8 - Comparing Quantities
• NCERT Solutions for Chapter 9 - Algebraic Expressions and Identities
• NCERT Solutions for Chapter 10 - Visualizing Solid Shapes
• NCERT Solutions for Chapter 11 - Mensuration
• NCERT Solutions for Chapter 12 - Exponents and Powers
• NCERT Solutions for Chapter 13 - Direct and Inverse Proportions
• NCERT Solutions for Chapter 14 - Factorization
• NCERT Solutions for Chapter 15 - Introduction to Graphs
• NCERT Solutions for Chapter 16 - Playing with Numbers

Why NCERT Solutions for Class 8 Maths by Studyrankers?

Class 8 is one of the important step in the life of every individual. The concepts present in this class will prove very beneficial in upcoming classes so you must clear your concepts. Maths Class 8 NCERT Solutions you will find here will be helpful for you in following ways:

• Get to know about the weaker portion in the syllabus.
• How to apply formulas easily and effectively
• By taking help of these NCERT Solutions for Class 8, you can easily solve supplementary textbooks such as RS Aggarwal, RD Sharma.
• You can boost your marks in the examinations.

NCERT Solutions will guide you in a better way to solve different problems which find you difficult. Class 8 Maths NCERT Solutions has total 16 chapters. The textbook start with Rational Numbers and ends with an engaging chapter named Playing with numbers. In between you will study number systems, finding squares and cube roots, geometry and graphs. These chapters will guide in solving a numerous numbers of questions in higher classes and also in various competitive examinations. Class 8 will play a pivotal role in shaping your career so you must have conceptual clarity of NCERT Solutions for Class 8 Maths.

Chapter 1 - Rational Numbers

Rational numbers can be expressed in the form of p/q, where q cannot be equal to zero. In this chapter we will dealing with real numbers, integers, whole numbers, natural numbers and properties of rational numbers such as independent, associative and closure. We will also learn about additive identity (0) and also the multiplicative identity (1). There are two exercises in the chapter which has 18 questions in total.

NCERT Solutions for Class 8 Maths Chapter 1 Exercises
Exercise 1.1 Class 8 Maths Solutions Exercise 1.2 Class 8 Maths Solutions

Chapter 2 - Linear Equations in One Variable

An equation is said to be linear when its highest power of the variable is 1. In this chapter, you will learn how to balance right hand side and left hand side of the equation. We will apply them on various word problems based on age, currency etc. This chapter has six exercises which are divided into 65 questions.

NCERT Solutions for Class 8 Maths Chapter 2 Exercises
Exercise 2.1 Class 8 Maths Solutions Exercise 2.2 Class 8 Maths Solutions Exercise 2.3 Class 8 Maths Solutions Exercise 2.4 Class 8 Maths Solutions Exercise 2.5 Class 8 Maths Solutions Exercise 2.6 Class 8 Maths Solutions

Chapter 3 - Understanding Quadrilaterals

Based on the number of sides or vertices, polygons can be classified as a triangle (3 sides), quadrilateral (4 sides), pentagon (5 sides), hexagon (6 sides) and so on. In this chapter, we will learn about the properties of different types of quadrilaterals such as parallelogram, rhombus, rectangle, square and kite. This chapter has four exercises in which there are 31 questions.

NCERT Solutions for Class 8 Maths Chapter 3 Exercises
Exercise 3.1 Class 8 Maths Solutions Exercise 3.2 Class 8 Maths Solutions Exercise 3.3 Class 8 Maths Solutions Exercise 3.4 Class 8 Maths Solutions

Chapter 4 - Practical Geometry

NCERT Solutions for Class 8 Maths Chapter 4 is about the construction of quadrilaterals. We will learn to construct quadrilaterals when Four sides and one diagonal are given, two diagonals and three sides are given, two adjacent sides and three angles are given, three sides and two included angles are given and when other special properties are known. There are total 5 exercises which which has 8 questions.

NCERT Solutions for Class 8 Maths Chapter 4 Exercises
Exercise 4.1 Class 8 Maths Solutions Exercise 4.2 Class 8 Maths Solutions Exercise 4.3 Class 8 Maths Solutions Exercise 4.4 Class 8 Maths Solutions Exercise 4.5 Class 8 Maths Solutions

Chapter 5 - Data Handling

The information generally numerically that are collected through observation are called Data. This chapter is about drawing meaningful interpretation through pictograph, histograms, pie charts, bar graphs, etc. The chapter has three exercises which has total 16 questions.

NCERT Solutions for Class 8 Maths Chapter 5 Exercises
Exercise 5.1 Class 8 Maths Solutions Exercise 5.2 Class 8 Maths Solutions Exercise 5.3 Class 8 Maths Solutions

Chapter 6 - Square and Square Roots

A number y whose square is x. Numbers that end with 0,1, 4, 5, 6 or 9 are called square numbers. The chapter is about the properties of square numbers, finding the square of a number, finding square roots through various methods and Square roots of decimals. The chapter has four exercises which has 30 questions in total.

NCERT Solutions for Class 8 Maths Chapter 6 Exercises
Exercise 6.1 Class 8 Maths Solutions Exercise 6.2 Class 8 Maths Solutions Exercise 6.3 Class 8 Maths Solutions Exercise 6.4 Class 8 Maths Solutions

Chapter 7 - Cube and Cube Roots

This chapter is about finding the cubes and cube roots of different numbers through various methods and which number is perfect cube. The chapter has two exercises which has total 7 questions.

NCERT Solutions for Class 8 Maths Chapter 7 Exercises
Exercise 7.1 Class 8 Maths Solutions Exercise 7.2 Class 8 Maths Solutions

Chapter 8 - Comparing Quantities

Chapter 8 Class 8 Maths NCERT Solutions mostly deals with word problems that are part of our daily lives. The questions are based on finding percentage, interest, profit and loss, discounts and goods and services tax etc. There are three exercises which has total 28 questions.

NCERT Solutions for Class 8 Maths Chapter 8 Exercises
Exercise 8.1 Class 8 Maths Solutions Exercise 8.2 Class 8 Maths Solutions Exercise 8.3 Class 8 Maths Solutions

Chapter 9 - Algebraic Expressions and Identities

An algebraic expression is an expression built up from integer constants, variables, and the algebraic operations. Expressions that contain exactly one, two and three terms are called monomials, binomials and trinomials, respectively. We will study various algebraic identities. There are 5 exercises in the whole chapter and total 25 questions.

NCERT Solutions for Class 8 Maths Chapter 9 Exercises
Exercise 9.1 Class 8 Maths Solutions Exercise 9.2 Class 8 Maths Solutions Exercise 9.3 Class 8 Maths Solutions Exercise 9.4 Class 8 Maths Solutions Exercise 9.5 Class 8 Maths Solutions

Chapter 10 - Visualizing Solid Shapes

You will learn about three-dimensional shapes in this chapter. We will talk about Euler’s formula which deals with the relationship between faces, vertices, and edges. There are three exercises in this chapter which consist of total 16 questions.

NCERT Solutions for Class 8 Maths Chapter 10 Exercises
Exercise 10.1 Class 8 Maths Solutions Exercise 10.2 Class 8 Maths Solutions Exercise 10.3 Class 8 Maths Solutions

Chapter 11 - Mensuration

We will focused on solving problems related to perimeter and areas of plane closed figure and surface area and volume of some solid figure like cube, cuboid, cylinder etc. The chapter has four exercises in which 34 questions are given.

NCERT Solutions for Class 8 Maths Chapter 11 Exercises
Exercise 11.1 Class 8 Maths Solutions Exercise 11.2 Class 8 Maths Solutions Exercise 11.3 Class 8 Maths Solutions Exercise 11.4 Class 8 Maths Solutions

Chapter 12 - Exponents and Powers

In this chapter, we will discuss about exponents, laws of exponents and use of Exponents to Express Small Numbers in Standard Form. There are only 2 exercises in this chapter in which 11 questions in total.

NCERT Solutions for Class 8 Maths Chapter 12 Exercises
Exercise 12.1 Class 8 Maths Solutions Exercise 12.2 Class 8 Maths Solutions

Chapter 13 - Direct and Inverse Proportions

Two variables x and y are said to be in direct proportion when the value of y increases with respect to increase in the value of x such that the ratio x/y does not change while two variables x and y are said to be in inverse proportion, when x increases y decreases and vice-versa. The 2 exercises in this chapter have 21 questions in total.

NCERT Solutions for Class 8 Maths Chapter 13 Exercises
Exercise 14.1 Class 8 Maths Solutions Exercise 14.2 Class 8 Maths Solutions

Chapter 14 - Factorization

Class 8 Maths NCERT Solutions Chapter 14 is about factorization of natural numbers and algebraic expression. You will learn factorisation using identities. There are 4 exercises in this chapter that consist of 34 questions.

NCERT Solutions for Class 8 Maths Chapter 14 Exercises
Exercise 14.1 Class 8 Maths Solutions Exercise 14.2 Class 8 Maths Solutions Exercise 14.3 Class 8 Maths Solutions Exercise 14.4 Class 8 Maths Solutions

Chapter 15 - Introduction to Graphs

In this chapter, you will learn graphical representation of data and about various types of graphs such as bar graphs, pie graphs, line graphs and linear graphs. We will be also be solving problems based on its application. There are three exercises which have 13 questions in total.

NCERT Solutions for Class 8 Maths Chapter 15 Exercises
Exercise 15.1 Class 8 Maths Solutions Exercise 15.2 Class 8 Maths Solutions Exercise 15.3 Class 8 Maths Solutions

Chapter 16 - Playing with Numbers

Its all about fun! You will find various games with numbers and puzzles. You will also learn divisibility test. There are 2 exercise in this chapter with 14 questions.

NCERT Solutions for Class 8 Maths Chapter 16 Exercises
Exercise 16.1 Class 8 Maths Solutions Exercise 16.2 Class 8 Maths Solutions

Why NCERT Solutions of StudyRankers?

Studyrankers have prepared the Class 8 Maths NCERT Solutions in such a way that you can approach the examinations easily and score good marks. You can access the NCERT Solutions from any device. Additionally, you can download Studyrankers app to access it offline without any internet connection. These Math Class 8 Solutions are provided free so every student can access it and prepare well for exams. You can conveniently access the study materials prepared by our experts like NCERT Solutions for Class 8 Maths, Class 8 Maths Notes, MCQ Questions for Class 8 Maths, Important Questions for Class 8 Maths. An individual can learn at their own pace through these and enhance their knowledge and problem solving skills.

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You can acces NCERT Solutions for Class 8 Maths just by visiting Studyrankers. These are prepared by experts who have great experience and deep knowledge of the subjects so you can trust on us. Every solutions is explained properly.

How many chapters are there in Class 8 NCERT Maths textbook?

There are total 16 chapter in the NCERT textbook for Class 8 Maths which will help you understanding the topics well. The chapters are Chapter 1 - Rational Numbers, Chapter 2 - Linear Equations in One Variable, Chapter 3 - Understanding Quadrilaterals, Chapter 4 - Practical Geometry, Chapter 5 - Data Handling, Chapter 6 - Squares and Square Roots, Chapter 7 - Cubes and Cube Roots, Chapter 8 - Comparing Quantities, Chapter 9 - Algebraic Expressions and Identities, Chapter 10 - Visualizing Solid Shapes, Chapter 11 - Mensuration, Chapter 12 - Exponents and Powers, Chapter 13 - Direct and Inverse Proportions, Chapter 14 - Factorization, Chapter 15 - Introduction to Graphs and Chapter 16 - Playing with Numbers.

How to download Class 8 Maths NCERT Solutions as PDF?

We have provided Maths Class 8 NCERT Solutions also as PDF which is free so you can study at your comfortability in your own anytime.

NCERT Solutions for Class 8 Maths Ch 13 Direct and Inverse Proportions Exercise 13.1

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NCERT Solutions for Class 8 Maths Ch 13 Direct and Inverse Proportions Exercise 13.1

Chapter 13 Direct and Inverse Proportions Exercise 13.1 NCERT Solutions for Class 8 Maths is helpful in understanding the key concepts inside the chapter properly. You can find NCERT Solutions here which will useful in clearing your doubts which you're finding very difficult to solve. Through the help of these NCERT Solutions for Class 8, one can clear their doubts and improve their learning behaviour.

Class 8 Maths NCERT Solutions is useful in understanding the important points inside the chapter properly. In exercise 13.1, you need to solve word problems.

NCERT Solutions for Class 8 Maths Ch 14 Factorization Exercise 14.4

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NCERT Solutions for Class 8 Maths Ch 14 Factorization Exercise 14.4

Chapter 14 Factorization Exercise 14.4 Class 8 Maths NCERT Solutions is very helpful in solving the difficult problems in a exercise. These NCERT Solutions are useful in clearing doubts and completing your homework on time and clear your doubts. You can also download our app Studyrankers for offline NCERT Solutions and learn efficiently.

NCERT Solutions for Class 8 Maths is a better guide to solve your queries whenever got stuck in a question. In Exercise 14.4, you have to find and correct the errors in the following mathematical statements.

NCERT Solutions for Class 8 Maths Ch 15 Introduction to Graphs Exercise 15.1

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NCERT Solutions for Class 8 Maths Ch 15 Introduction to Graphs Exercise 15.1

Chapter 15 Introduction to Graphs Geometry Exercise 15.1 NCERT Solutions for Class 8 Maths is provided here which are very beneficial in understanding the concepts inside the topic properly. These NCERT Solutions are prepared by Studyrankers experts which are detailed and accurate. Class 8 Maths will help you in gaining good marks in the examinations.

In Exercise 15.1, you have to study the graphs and answer the questions given below.

NCERT Solutions for Class 7 Maths Ch 6 The Triangles and its properties

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NCERT Solutions for Class 7 Maths Ch 6 The Triangles and its properties

Here you will find NCERT Solutions for Class 7 Maths Ch 6 The Triangles and its properties which will be helpful in knowing the important points and formulas inside the chapter. It will be useful in passing examinations with flying colours. These NCERT Solutions for Class 7 are accurate and detailed that will increase concentration among students.

1. In APQR, D is the mid-point of 

 is ________

PD is_______

Is QM = MR?

Answer

QD = DR

∴  is altitude.

PD is median.

No, QM (eq) MR as D is the mid-point of QR.

2. Draw rough sketches for the following: 

(a) In ΔABC, BE is a median. 

(b) In ΔPQR, PQ and PR are altitudes of the triangle. 

(c) In ΔXYZ, YL is an altitude in the exterior of the triangle.

Answer

(a) Here, BE is a median in ΔABC and AE = EC.

(b) Here, PQ and PR are the altitudes of the ΔPQR and RP ⊥ QP.

(c) YL is an altitude in the exterior of ΔXYZ.

3. Verify by drawing a diagram if the median and altitude of a isosceles triangle can be same. 

Answer

Isosceles triangle means any two sides are same.

Take ΔABC and draw the median when AB = AC.

AL is the median and altitude of the given triangle.

Exercise 6.2

1. Find the value of the unknown exterior angle x in the following diagrams:

Answer

(i) x = 50° + 70° = 120°

(ii) x = 65° + 45° = 110°

(iii) x = 30° + 40° = 70°

(iv) x = 60°+60° = 120°

(v) x = 50° + 50° = 100°

(vi) x = 60° + 30° = 90°

2. Find the value of the unknown interior angle x in the following figures:

Answer

(i)  x + 50° = 115°   ⇒    x = 115° - 50° = 65°

(ii)  70°+ x = 100°    ⇒   x = 100°- 70° =  30°

(iii)  x + 90° = 125°  ⇒  x = 120°- 90° =  35°

(iv)  60°+ x = 120°   ⇒  x = 120°- 60° = 60°

(v)   30° + x = 80°   ⇒   x = 80°- 30° = 50°

(vi)  x + 35°= 75°   ⇒ x = 75°- 35° = 40°

Exercise 6.3

1. Find the value of unknown x in the following diagrams:

Answer

(i) In ΔABC,

∠ BAC + ∠ ACB + ∠ ABC = 180°  [ By angle sum property of a triangle]

⇒ x + 50°+ 60° = 180°

⇒  x  + 110° = 180°

⇒  x = 180°-110° = 70°

(ii) In ΔPQR,

∠ RPQ + ∠ PQR + ∠ RPQ = 180° [By angle sum property of a triangle]

⇒  90°+30°+ x = 180°

⇒  x+ 120° = 180°

⇒   x= 180°-120°= 60°

(iii) In ΔXYZ,

∠ ZXY + ∠ XYZ + ∠ YZX = 180° [By angle sum property of a triangle]

⇒   30° + 110° + x = 180°

⇒  x + 140° = 180°

⇒ x = 180°-140° = 40°

(iv)  In the given isosceles triangle,

x+x + 50° = 180°   [By angle sum property of a triangle]

⇒  2x+50°= 180°

⇒  2x = 180°- 50°

⇒  2x = 130°

⇒ x = 130°/2 = 65°

(v) In the given equilateral triangle,

x +x+x = 180° [By angle sum property of a triangle]

⇒ 3x = 180°

⇒ x = 180°/3 = 60°

(vi) In the given right angled triangle,

x + 2x+90° = 180°    [By angle sum property of a triangle)

⇒  3x+90° = 180°

⇒  3* = 180°- 90°

⇒ 3x = 90°

⇒ x = 90°/3 = 30°

2. Find the values of the unknowns x and y in the following diagrams:

Answer

(i) 50° + x = 120°    [Exterior angle property of a Δ]

⇒   x = 120°-50° = 70°

Now, 50° + x + y =180°    [Angle sum property of a Δ]

⇒  50° + 70° + y = 180°

⇒  120° + y= 180°

⇒  y = 180° -120° = 60°

(ii) y = 80° ..... (i)    [Vertically opposite angle]

Now, 50° + x + y =180°    [Angle sum property of a Δ]

⇒  50° + 80°+y = 180°    [From equation (i)]

⇒ 130° + y = 180°

⇒ y = 180° -130° = 50°

(iii) 50°+ 60° = x    (Exterior angle property of a Δ]

x = 110°

Now 50° + 60°+ y = 180°    [Angle sum property of a Δ]

⇒  110° + y = 180°

⇒  y = 180° - 110°

⇒  y = 70°

(iv) x = 60° ..... (i) [Vertically opposite angle]

Now, 30° + x + y = 180°    [Angle sum property of a Δ ]

⇒  50° + 60° + y = 180°    [From equation (i)]

⇒ 90° + y = 180°

⇒  y = 180° - 90° = 90°

(v) y = 90°   ....(i)    [Vertically opposite angle]

Now, y + x + x = 180°    [Angle sum property of a Δ]

⇒  90° + 2x = 180°    [From equation (i)]

⇒ 2x = 180°- 90°

⇒  2x = 90°

⇒ x = 90°/2 = 45°

(vi) x = y  ...(i)   [Vertically opposite angle]

Now, x + x + y = 180°    [Angle sum property of a Δ]

⇒ 2x+x = 180°    [From equation (i)]

⇒ 3x = 180°

⇒ x = 180°/3 = 60°

Exercise 6.4

1. Is it possible to have a triangle with the following sides? 

(i) 2 cm, 3 cm, 5 cm 

(ii) 3 cm, 6 cm, 7 cm 

(iii) 6 cm, 3 cm, 2 cm 

Answer

(i) 2 cm, 3 cm, 5 cm

2 + 3 > 5 No

2 + 5 > 3 Yes

3 + 5 > 2 Yes

This triangle is not possible.

(ii) 3 cm, 6 cm, 7 cm

3 + 6 > 7 Yes

6 + 7 > 3 Yes

3 + 7 > 6 Yes

This triangle is possible.

(iii) 6 cm, 3 cm, 2 cm

6 + 3 > 2 Yes

6 + 2 > 3 Yes

2 + 3 > 6 No

This triangle is not possible.

2. Take any point O in the interior of a triangle PQR. Is:

(i) OP + OQ > PQ ?

(ii) OQ + OR > QR?

(iii) OR + OP > RP?

Answer

Join OR, OQ and OP.

(i) Is OP + OQ > PQ ?

Yes, POQ form a triangle.

(ii) Is OQ + OR > QR ?

Yes, RQO form a triangle.

(iii) Is OR + OP > RP ?

Yes, ROP form a triangle.

3. AM is a median of a triangle ABC. Is AB + BC + CA > 2 AM? (Consider the sides of triangles ΔABM and ΔAMC.)

Answer

Since, the sum of lengths of any two sides in a triangle should be greater titan the length of third side.

Therefore, In ΔABM, AB + BM > AM ... (i)

In ΔAMC, AC + MC > AM ... (ii)

Adding eq. (i) and (ii),

AB + BM + AC + MC > AM + AM 

⇒ AB + AC + (BM + MC) > 2AM 

⇒ AB + AC + BC > 2AM

Hence, it is true.

4. ABCD is a quadrilateral. Is AB + BC + CD + DA > AC + BD?

Answer

Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side.

Therefore, In Δ ABC, AB + BC > AC .........(i)

In Δ ADC, AD + DC > AC  (ii)

In ΔDCB, DC + CB > DB  (iii)

In ΔADB, AD + AB > DB   (iv)

Adding equations (i), (ii), (iii) and (iv), we get

AB + BC + AD + DC + DC + CB + AD + AB > AC + AC + DB + DB 

⇒ (AB + AB) + (BC + BC) + (AD + AD) + (DC + DC) > 2AC + 2DB 

⇒ 2AB + 2BC +2AD + 2DC > 2(AC+DB)

⇒ 2(AB +BC + AD +DC) > 2(AC +DB)

⇒ AB + BC + AD + DC > AC + DB

⇒ AB + BC + CD + DA > AC + DB

Hence, it is true.

5. ABCD is quadrilateral. Is AB + BC + CD + DA < 2 (AC + BD)?

Answer

Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side.

Therefore, In A AOB, AB < OA + OB ......... (i)

In A BOC, BC < OB + OC  (ii)

In A COD, CD<OC + OD  (iii)

InAAOD, DA < OD + OA  ....(iv)

Adding equations (i), (ii), (iii) and (iv), we get

AB + BC + CD + DA < OA + OB + OB + OC + OC + OD + OD + OA 

⇒ AB + BC + CD + DA < 2OA + 20B + 2OC + 2OD

⇒ AB + BC + CD + DA < 2[(AO + OC) + (DO + OB)]

⇒ AB + BC + CD + DA < 2(AC + BD)

Hence, it is proved.

6. The lengths of two sides of a triangle are 12 cm and 15 cm. Between what two measures should the length of the third side fall? 

Answer

Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side.

It is given that two sides of triangle are 12 cm and 15 cm.

Therefore, the third side should be less than 12 + 15 = 27 cm.

Also the third side cannot be less than the difference of the two sides.

Therefore, the third side has to be more than 15 – 12 = 3 cm. Hence, the third side could be the length more than 3 cm and less than 27 cm.

Exercise 6.5

1. PQR is a triangle, right angled at P. If PQ = 10 cm and PR = 24 cm, find QR.

Answer

Given: PQ = 10 cm, PR = 24 cm

Let QR be x cm.

In right angled triangle QPR,

[Hypotenuse)² = (Base)² + (Perpendicular)²   [By Pythagoras theorem]

⇒ (QR)² = (PQ)² + (PR]²

⇒ x² = (10)² +(24)² 

⇒ x² = 100 + 576 = 676

= 26 cm

Thus, the length of QR is 26 cm.

2. ABC is a triangle, right angled at C. If AB = 25 cm and AC = 7 cm, find BC.

Answer

Given: AB = 25 cm, AC = 7 cm

 Let BC be x cm.

 In right angled triangle ACB,

 (Hypotenuse)2 = (Base)2 + (Perpendicular)2      [By Pythagoras theorem]

 ⇒ (AB)2 = (AC)2 + (BC)2

 ⇒ (25)2 = (7)2+x2

 ⇒ 625 = 49 + x2

 ⇒ x2 = 625 - 49 = 576

= 24 cm

Thus, the length of BC is 24 cm.

3. A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.

Answer

Let AC be the ladder and A be the window.

Given: AC = 15 m, AB = 12 m, CB = a m

In right angled triangle ACB,

(Hypotenuse)² = (Base)² + (Perpendicular)²    [By Pythagoras theorem]

⇒ (AC)² = (CB)² + (AB)²

⇒ (15)² + (a)² = (12)²

⇒ 225 = a² + 144

⇒ a² = 225 - 144 = 81

= 9 m

Thus, the distance of the foot of the ladder from the wall is 9 m.

4. Which of the following can be the sides of a right triangle? 

(i) 2.5 cm, 6.5 cm, 6 cm 

(ii) 2 cm, 2 cm, 5 cm 

(iii) 1.5 cm, 2 cm, 2.5 cm 

In the case of right angled triangles, identify the right angles.

Answer

Let us consider, the larger side be the hypotenuse and also using Pythagoras theorem,

(Hypotenuse)² = (Base)² + (Perpendicular)²

(i) 2.5 cm, 6.5 cm, 6 cm

In AABC, (AC)² = (AB)² + (BC)²

L.H.S. = (6.5)² = 42.25 cm

R.H.S. = (6)² + (2.5)² = 36 + 6.25 = 42.25 cm

Since, L.H.S. = R.H.S.

Therefore, the given sides are of the right angled triangle.

Right angle lies on the opposite to the greater side 6.5 cm, i.e., at B.

(ii) 2 cm, 2 cm, 5 cm

In the given triangle, (5)² = (2)² + (2)² 

L.H.S. = (5)² = 25

R.H.S. = (2)² + (2)² = 4 + 4 = 8 

Since, L.H.S. ≠ R.H.S.

Therefore, the given sides are not of the right angled triangle.

(iii) 1.5 cm, 2 cm, 2.5 cm

In ΔPQR, (PR)² = (PQ)² + (RQ)²

L.H.S. = (2 5)² = 6.25 cm

R.H.S. = (1.5)² + (2)² = 2.25 + 4 = 6.25 cm

Since, L.H.S. = R.H.S.

Therefore, the given sides are of the right angled triangle.

Right angle lies on the opposite to the greater side 2.5 cm, i.e., at Q.

5. A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree. 

Answer

Let A'CB represents the tree before it broken at the point C and let the top A’ touches the ground at A after it broke. Then ΔABC is a right angled triangle, right angled at B.

AB = 12 m and BC = 5 m

Using Pythagoras theorem, In ΔABC

(AC)² =(AB)² + (BC)² 

⇒ (AC)² = (12)² + (5)²

⇒ (AC)² = 144 + 25

⇒ (AC)² = 169

⇒ AC = 13 m

Hence, the total height o f the tree = AC + CB 13 + 5 = 18 m.

6. Angles Q and R of a ΔPQR are 25° and 65°

Write which of the following is true:

(i) PQ² + QR² = RP²

(ii) PQ² + RP² = QR² 

(iii) RP² + QR² = PQ²

Answer

In ΔPQR, ∠PQR + ∠QRP + ∠RPQ = 180°      [By Angle sum property of a Δ] 

⇒ 25° + 65° + ∠RPQ = 180°

⇒ 90° + ∠RPQ = 180°

⇒ ∠RPQ = 180°- 90° = 90°

Thus, ΔPQR is a right angled triangle, right angled at P.

∴ (Hypotenuse)² = (Base)² + (Perpendicular)²    [By Pythagoras theorem]

⇒ (QR)² = (PR)² + (QP)²

Hence, Option (ii) is correct.

7. Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm. 

Answer

Given diagonal (PR) = 41 cm, length (PQ) = 40 cm

Let breadth (QR) be x cm.

Now, in right angled triangle PQR,

(PR)² = (RQ)² +(PQ)²      [By Pythagoras theorem]

⇒ (41)² = x² + (40)²

⇒ 1681 = x² + 1600

⇒ x² = 1681 - 1600

⇒ x² = 81

= 9 cm

Therefore the breadth of the rectangle is 9 cm,

Perimeter of rectangle - 2(length + breadth)

= 2 [9 + 49)

= 2 × 49 = 98 cm

Hence, the perimeter of the rectangle is 98 cm.

8. The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter. 

Answer

Given: Diagonals AC = 30 cm and DB = 16 cm.

Since the diagonals of the rhombus bisect at right angle to each other.

Therefore, OD = DB/2 = 16/2 = 8 cm

And OC = AC/2 = 30/2 = 15 cm

Now , In right angle triangle DOC,

(DC)2 =(OD)2 + (OC)2    [By Pythagoras dieorem]

⇒  (DC)2 = (8)2+(15)2

⇒ (DC)2 = 64 + 225 = 289

= 17 cm

Perimeter of rhombus = 4 x side = 4 x 17 = 68 cm

Thus, die perimeter of rhombus is 68 cm.

---

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Class 8 Maths NCERT Solutions is useful in understanding the important points inside the chapter properly. In Exercise 16.1, you have to find the values of the letters.

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You can also download our app Studyrankers for offline study materials. In exercise 16.2, you have to find the values of the letters.

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MCQ Questions for Class 11 Sociology: Ch 1 Social Structure, Stratification and Social Processes

1. The term 'social structure' means that society

(a) is stratified.

(b) has concrete structure.

(c) is captured.

(d) is structured.

► (a) is stratified.

2. The notion that ‘human beings have to cooperate to meet their basic needs, and to produce and reproduce themselves and their world’ was supported by

(a) Emile Durkheim.

(b) Charles Darwin.

(c) Tom Bottomore.

(d) Durkheim and Karl Marx.

► (d) Durkheim and Karl Marx.

3. Admission procedures, codes of conduct, daily assemblies, annual functions and school anthems are the examples of

(a) family structure.

(b) political structure.

(c) school structure.

(d) religious structure.

► (c) school structure.

4. The universal features of all societies are

(a) growth, development and progress.

(b) co-operation, conflict and competition.

(c) urbanisation and industrialisation.

(d) tradition and modernity.

► (b) co-operation, conflict and competition.

5. 'The clash of interests' is known as

(a) conflict.

(b) competition.

(c) co-operation.

(d) class struggle.

► (a) conflict.

6. ‘Expansion of trade, division of labour, specialisation and rising productivity’ are the features of

(a) post-industrialisation.

(b) democracy.

(c) capitalism. 

(d) religious rituals.

► (c) capitalism. 

7. According to Emile Durkheim, societies exert constraint over

(a) social structure.

(b) social institutions.

(c) social groups.

(d) the actions of their members.

► (d) the actions of their members.

8. The form of social cohesion that is based on division of labour is known as

(a) social bond.

(b) social solidarity.

(c) social unity.

(d) organic solidarity.

► (d) organic solidarity.

9. With co-operation, conflict and competition are also the parts of

(a) modern societies.

(b) tribal societies.

(c) rural societies.

(d) Indian societies.

► (a) modern societies.

10. Unit of social structure is

(a) institution.

(b) association.

(c) social groups.

(d) all of these.

► (d) all of these.

11. “Social structure constrains our activities to constructed frameworks and patterns of organisation and directs our behaviour” Name the thinker

(a) Max Weber

(b) Durkheim

(c) Bottomore

(d) K. Marx

► (b) Durkheim

12. Life Chances, Social Statuses and political influences are enjoyed by  __________.

(a) Primary Groups

(b) Tertiary Groups

(c) Privilege Groups

(d) Interest Groups

► (c) Privilege Groups

13. The choices an individual makes/has in life, in terms of the school s/he goes to , or if s/he goes to school at all, the clothes s/he gets to wear, the foods /he consumes etc all of which is determined by _______.

(a) Status

(b) Role

(c) Prestige

(d) All the above.

► (a) Status

14. The central concepts to understanding the dialectical relationship between the society and individual are _______.

(a) Structure

(b) Stratification

(c) Social processes

(d) All the above

► (d) All the above

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Chapter 1 Integers Exercise 1.1 NCERT Solutions for Class 7 Maths is very helpful regarding the preparation of examinations. These NCERT Solutions for Class 7 are detailed and accurate through which you can clear your doubts. You will get to now about a lot of important points given in the chapter. NCERT Solutions for Class 7 Maths is best way through which a student can resolve their queries and able to solve the difficult problems in a given exercise. In exercise 1.1, you have to do calculation and answers question by looking into picture.

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Exercise 7.1

1. Complete the following statements: 

(a) Two line segments are congruent if _______________. 

(b) Among two congruent angles, one has a measure of 70o, the measure of other angle is __________. 

(c) When we write ∠A = ∠B, we actually mean _______________. 

Answer

(a) they have the same length

(b) 70°

(c) m∠A = m∠B

2. Give any two real time examples for congruent shapes. 

Answer

(i) Two footballs

(ii) Two teacher’s tables

3. If ΔABC ≌  ΔFED under the correspondence ABC↔FED, write all the corresponding congruent parts of the triangles.

Answer

Given: ΔABC ≌  ΔFED.

The corresponding congruent parts of die triangles are:

(i) ∠A ↔ ∠F

(ii) ∠B ↔ ∠E

(iii) ∠C ↔ ∠D

4. If ΔDEF ≌  ΔBCA, write the part (s) of ΔBCA that correspond to:

Answer

Exercise 7.2 

1. Which congruence criterion do you use in the following? 

(a) Given: AC = DF, AB = DE, BC = EF

So ΔABC ≌ ΔDEF

(b) Given: RP = ZX, RQ = ZY, ∠PRQ = ∠XZY
So ΔPQR ≌ ΔXYZ

(c) Given: ∠ MLN = ∠ FGH, ∠ NML = ∠ HFG, ML = FG
So ΔLMN ≌ ΔGFH

(d) Given: EB = BD, AE = CB, ∠A = ∠C = 90°
So AABE = ACDB

Answer

(a) By SSS congruence criterion, since it is given that AC = DF, AB = DE, BC = EF
The three sides of one triangle are equal to the three corresponding sides of another triangle.
Therefore, ΔABC ≌ ΔDEF

(b) By SAS congruence criterion, since it is given that RP = ZX, RQ = ZY and ∠PRQ = ∠XZY
The two sides and one angle in one of the triangle are equal to the corresponding sides and the angle of other triangle.
Therefore, ΔPQR ≌ ΔXYZ

(c) By ASA congruence criterion, since it is given that ∠MLN = ∠FGH, ∠NML = ∠HFG, ML = FG.
The two angles and one side in one of the triangle are equal to the corresponding angles and side of other triangle.
Therefore, ΔLMN ≌ ΔGFH

(d) By RHS congruence criterion, since it is given that EB = BD, AE = CB, ∠A = ∠C = 90°
Hypotenuse and one side of a right angled triangle are respectively equal to the hypotenuse and one side of another right angled triangle.
Therefore, ΔABE ≌ ΔCDB.

2. You want to show that ΔART ≌ ΔPEN:
(a) If you have to use SSS criterion, then you need to show:
(i) AR =
(ii) RT =
(iii) AT =

(b) If it is given that ∠T = ∠N and you are to use SAS criterion, you need to have:
(i) RT = and
(ii) PN =

(c) If it is given that AT = PN and you are to use ASA criterion, you need to have:
(i) ?
(ii) ?

Answer

(a) Using SSS criterion, ΔART ≌ ΔPEN
(i) AR = PE
(ii) RT= EN
(iii) AT = PN

(b) Given: ∠T = ∠N
Using SAS criterion, ΔART ≌ ΔPEN
(i) RT = EN
(ii) PN = AT

(c) Given: AT = PN
Using ASA criterion, ΔART ≌ ΔPEN
(i) ∠RAT = ∠EPN
(ii) ∠RTA = ∠ENP

3. You have to show that ΔAMP = ΔAMQ. In the following proof supply the missing reasons:

Steps	Reasons
(i) PM = QM	(i)
(ii) ∠ PMA = ∠ QMA	(ii)
(iii) AM = AM	(iii)
(iv) ΔAMP ≌ ΔAMQ	(iv)

Answer

Steps	Reasons
(i) PM = QM	(i) Given
(ii) ∠ PMA = ∠ QMA	(ii) Given
(iii) AM = AM	(iii) Common
(iv) ΔAMP ≌ ΔAMQ	(iv) SAS congruence rule

4. In ΔABC, ∠A = 30° ∠B = 40° and ∠C = 110°

In ΔPQR, ∠P = 30° ∠Q = 40° and ∠R = 110°.

A student says that ΔABC ≌ ΔPQR by AAA congruence criterion. Is he justified? Why or why not?

Answer

No, because the two triangles with equal corresponding angles need not be congruent. In such a correspondence, one of them can be an enlarged copy of the other.

5. In the figure, the two triangles are congruent. The corresponding parts are marked. We can write Δ RAT ≌ ?

Answer

In the figure, given two triangles are congruent. So, the corresponding parts are:

A↔O

R↔W

T↔N

We can write, ΔRAT ≌ ΔWON    [By SAS congruence rule]

6. Complete the congruence statement:

Answer

In A BAT and ABAC, given triangles are congruent so the corresponding parts are:

B↔B

A↔A

T↔C

Thus, ΔBCA ≌ ΔBTA    |By SSS congruence rule]

In ΔQRS and ΔTPQ, given triangles are congruent so the corresponding parts are:

P↔R

T↔Q

Q↔S

Thus, ΔQRS ≌  ΔTPQ    [By SSS congruence rule]

7. In a squared sheet, draw two triangles of equal area such that: 

(i) the triangles are congruent. 

(ii) the triangles are not congruent. 

What can you say about their perimeters? 

Answer

In a squared sheet, draw ΔABC and ΔPQR. When two triangles have equal areas and

(i)

In the above figure, ΔABC and ΔDEF have equal areas.

And also, ΔABC≌ΔDEF

So, we can say that perimeters of ΔABC and ΔDEF are equal.

(ii)

In the above figure, ΔLMN and ΔOPQ

ΔLMN is not congruent to ΔOPQ

So, we can also say that their perimeters are not same.

8. Draw a rough sketch of two triangles such that they have five pairs of congruent parts but still the triangles are not congruent. 

Answer

Let us draw two triangles PQR and ABC.

All angles are equal, two sides are equal except one side. Hence, ΔPQR are not congruent to ΔABC.

9. If ΔABC and ΔPQR are to be congruent, name one additional pair of corresponding parts. What criterion did you use?

Answer

A ABC and A PQR are congruent Then one additional pair is 

Given: ∠B = ∠Q = 90°

∠C/BC = ∠R/QR

Therefore, ΔABC ≌ ΔPQR    [By ASA congruence rule]

10. Explain, why ΔABC ≌ ΔFED.

Answer

Given: ∠A = ∠F, BC = ED, ∠B = ∠E
 In ΔABC and ΔFED,

 ∠B = ∠E = 90°

 ∠A = ∠F

 BC = ED

 Therefore, ΔABC ≌ ΔFED    [By RHS congruence rule]

---

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Exercise 8.1

1. Find the ratio of:
(a) Rs5 to 50 paise
(b) 15 kg to 210 g
(c) 9 m to 27 cm
(d) 30 days to 36 hours

Answer

To find ratios, both quantities should be in same unit.

(a) Rs 5 to 50 paise
⇒ 5 x 100 paise to 50 paise [∴ Rs 1 = 100 paise]
⇒ 500 paise to 50 paise

Thus, the ratio is = 500/50 = 10/1 = 10 : 1

(b) 15 kg to 210 g
⇒ 15 x 1000 g to 210 g [∵ 1kg = 1000 g]
⇒ 15000 g to 210 g
Thus, the ratio is = 15000/210 = 500/7 = 500:7

(c) 9 m to 27 cm
⇒ 9 x 100 cm to 27 cm [∵ 1m = 100 cm]
⇒ 900 cm to 27 cm

Thus, the ratio is = 900/27 = 100/3 = 100:3

(d) 30 days to 36 hours
⇒ 30 x 24 hours to 36 hours [∵ 1 day = 24 hours]
⇒ 720 hours to 36 hours

Thus, the ratio is = 720/36 = 20/1 = 20:1

2. In a computer lab, there are 3 computers for every 6 students. How many computers will be needed for 24 students? 

Answer

∵ 6 students need = 3 computers

∵ 1 student needs = 3/6 computers

∵ 24 students need = 

Thus, 12 computers will be needed for 24 students.

3. Population of Rajasthan = 570 lakhs and population of U.P. = 1660 lakhs. Area of Rajasthan = 3 lakh km² and area of U.P. = 2 lakh km². 

(i) How many people are there per km² in both states? 

(ii) Which state is less populated? 

Answer

(i) People present per km² = Population/Area

In Rajasthan = 570 lakhs/3 lakhs per km² = 190 people km2

In UP = 1660 lakhs/2 lakhs per km² = 830 people per km2

(ii) Rajasthan is less populated.

Exercise 8.2

1. Convert the given fractional numbers to percent: 

(a) 1/8

(b) 5/4

(c) 3/40

(d) 2/7

Answer

2. Convert the given decimal fractions to per cents: 

(a) 0.65

(b) 2.1

(c) 0.02

(d) 12.35

Answer

3. Estimate what part of the figures is coloured and hence find the percent which is coloured.

Answer

(i) Coloured part = 1/4

∴ Percent of coloured part 

(ii) Coloured part = 3/5

∴ Percent of coloured part 

(iii) Coloured part = 3/8

∴ Percent of coloured part = 37.5%

4. Find: 

(a) 15% of 250 

(b) 1% of 1 hour 

(c) 20% of 2500 

(d) 75% of 1 kg 

Answer

(a) 15% of 250 

(b) 1% of 1 hours = 1% of 60 minutes = 1% of (60 x 60] seconds

(c) 20% of Rs 500 

5. Find the whole quantity if: 

(a) 5% of it is 600 

(b) 12% of it is ₹1080

(c) 40% of it is 500 km 

(d) 70% of it is 14 minutes 

(e) 8% of it is 40 litres

Answer

Let the whole quantity be x in given questions:

6. Convert given per cents to decimal fractions and also to fractions in simplest forms:

(a) 25% 

(b) 150% 

(c) 20% 

(d) 5% 

Answer

S. No.	Percents	Fractions	Simplest form	Decimal form
(a)	25%	25/100	1/4	0.25
(b)	150%	150/100	3/2	1.5
(c)	20%	20/100	1/5	0.2
(d)	5%	5/100	1/20	0.05

7. In a city, 30% are females, 40% are males and remaining are children. What percent are children?

Answer

Given: Percentage of females =30%

Percentage of males = 40%

Total percentage of females and males = 30 + 40 = 70%

Percentage of children = Total percentage - Percentage of males and females

= 100%-70%

= 30%

Hence, 30% are children.

8. Out of 15,000 voters in a constituency, 60% voted. Find the percentage of voters who did not vote. Can you now find how many actually did not vote? 

Answer

Total voters = 15,000

Percentage of voted candidates = 60%

Percentage of not voted candidates = 100 - 60 = 40%

Actual candidates, who did not vote = 40% of 15000 

Hence, 6,000 candidates did not vote.

9. Meeta saves Rs 400 from her salary. If this is 10% of her salary. What is her salary?

Answer

Let Meera's salary be Rs x.

Now, 10% of salary = Rs 400

⇒ 10% of x = Rs 400

Hence, Meera's salary is Rs 4,000.

10. A local cricket team played 20 matches in one season. It won 25% of them. How many matches did they win? 

Answer

Number of matches played by cricket team = 20

Percentage of won matches = 25%

Total matches won by them = 25% of 20

= 5

Hence, they won 5 matches.

Exercise 8.3 

1. Tell what is the profit or loss in the following transactions. Also find profit percent or loss percent in each case. 

(a) Gardening shears bought for Rs 250 and sold for Rs 325. 

(b) A refrigerator bought Rs12,000 and sold at Rs 13,500. 

(c) A cupboard bought for Rs 2,500 and sold at Rs 3,000. 

(d) A skirt bought for Rs 250 and sold at Rs 150. 

Answer

(a) Cost price of gardening shears = Rs 250

Selling price of gardening shears = Rs 325

Since, S.P. > C.P., therefore here is profit.

∴ Profit = S.P. - C.P. = Rs 325 - Rs 250 = Rs 75 

Now,

(b) Cost price of refrigerator = Rs 12,000

Selling price of refrigerator = Rs 13,500

Since, S.P. > C.P., therefore here is profit.

∴ Profit = S.P. - C.P. = Rs 13500 - Rs 12000 = Rs 1,500

Now,

Therefore, Profit = Rs 1,500 and Profit% = 12.5%

(c) Cost price of cupboard = Rs 2,500

Selling price of cupboard = Rs 3,000

Since, S.P. > C.P., therefore here is profit.

∴ Profit = S.P. - CP. = Rs 3,000 - Rs 2,500 = Rs 500

Now,

Therefore, Profit = Rs 500 and Profit% = 20%

(d) Cost price of skirt = Rs 250

Selling price of skirt = Rs 150

Since, C.P. > S.P., therefore here is loss.

∴ Loss = CP, - S.P. = Rs 250 - Rs 150 = Rs100

Now,

Therefore, Profit = Rs 100 and Profit% = 40%

2. Convert each part of the ratio to percentage: 

(a) 3 : 1 

(b) 2 : 3 : 5 

(c) 1 : 4 

(d) 1 : 2 : 5 

Answer

(a) 3 :1 

Total part =3 + 1= 4

Therefore, Fractional part = 3/4 : 1/4

⇒ Percentage of parts 

⇒ Percentage of parts = 75% : 25%

(b) 2 : 3 : 5

Total part = 2 + 3 + 5 = 10

Therefore, Fractional part 

⇒ Percentage of parts 

⇒ Percentage of parts = 20% : 30% : 50%

(c) 1:4

Total part =1 + 4 = 5

Therefore, Fractional part = 1/5 : 4/5

⇒ Percentage of parts 

⇒ Percentage of parts = 20%: 00%

(d) 1 : 2 : 5

Total part = 1+ 2 + 5 = 8

Therefore, Fractional part 

⇒ Percentage of parts 

⇒ Percentage of parts = 12.5% : 25% : 62.5%

3. The population of a city decreased from 25,000 to 24,500. Find the percentage decrease.

Answer

The decreased population of a city from 25,000 to 24,500.

Population decreased = 25,000 - 24,500 = 500

Decreased Percentage 

Hence, the percentage decreased is 2%.

4. Arun bought a car for Rs 3,50,000. The next year, the price went up to Rs 3,70,000. What was the percentage of price increase?

Answer

Increased in price of a car from Rs 3,50,000 to Rs 3,70,000.

Amount change = Rs 3,70,000 - Rs 3,50,000 = Rs 20,000.

Therefore, Increased percentage 

Hence, the percentage of price increased is 

5. I buy a T.V. for Rs 10,000 and sell it at a profit of 20%. How much money do I get for it?

Answer

The cost price of T.V. = Rs 10,000

Profit percent = 20% Now,

Profit = Profit% of C.P, 

Selling price = C.P. + Profit = Rs 10,000 + Rs 2,000 = Rs 12,000

Hence, he gets Rs 12,000 on selling his T.V.

6. Juhi sells a washing machine for Rs 13,500. She loses 20% in the bargain. What was the price at which she bought it? 

Answer

Selling price of washing machine = Rs. 13,500

Loss percent = 20%

Let the cost price of washing machine be Rs x.

Since, Loss = Loss% of C.P.

⇒ 

Therefore, S.P. = C.P, - Loss

7. (i) Chalk contains Calcium, Carbon and Oxygen in the ratio 10:3:12. Find the percentage of Carbon in chalk. 

(ii) If in a stick of chalk, Carbon is 3 g, what is the weight of the chalk stick?

Answer

(i) Given ratio = 10 : 3 : 12

Total part = 10 + 3 + 12 = 25 

Part of Carbon = 3/25

Percentage of Carbon part in chalk 

(ii) Quantity of Carbon in chalk stick = 3 g

Let the weight of chalk be x g.

Then, 12 % of x = 3

Hence, the weight of chalk stick is 25 g.

8. Amina buys a book for Rs 275 and sells it at a loss of 15%. How much does she sell it for?

Answer

The cost of a book = Rs 275

Loss percent = 15%

Loss = Loss% of C.P. = 1 5 % of Rs 275

Therefore, S.P. = C.P. - Loss = Rs 275 - Rs 41.25 = Rs 233.75

Hence, Amina sells a book for Rs 233.75.

9. Find the amount to be paid at the end of 3 years in each case: 

(a) Principal = Rs 1,200 at 12% p.a. 

(b) Principal = Rs 7,500 at 5% p.a.

Answer

(a) Here, Principal (P) = Rs 1,200, Rate (R) = 12% p.a., Time (T) = 3 years

Now, Amount = Principal + Simple Interest

= Rs 1200 + Rs 432 

= Rs 1,632

(b) Here, Principal (P) = Rs 7,500, Rate (R) = 5% p.a., Time (T) = 3 years

Now, Amount = Principal + Simple Interest

= Rs 7,500 + Rs 1,125

= Rs 8,625

10. What rate gives Rs 280 as interest on a sum of Rs 56,000 in 2 years? 

Answer

Here, Principal (P) = Rs 56,000, Simple Interest (S.I.) = Rs 280, Time (T) = 2 years

Hence, the rate of interest on sum is 0.25%.

11. If Meena gives an interest of Rs 45 for one year at 9% rate p.a. What is the sum she has borrowed?

Answer

Simple Interest = Rs 45, Rate (R) = 9% p.a., Time (T) = 1 years

Hence, she borrowed Rs 500.

NCERT Solutions for Class 7 Maths Ch 1 Integers Exercise 1.3

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NCERT Solutions for Class 7 Maths Ch 1 Integers Exercise 1.3

Chapter 1 Integers Exercise 1.3 Class 7 Maths NCERT Solutions can be used to complete homework on time.  These NCERT Solutions for Class 7 will help you in grasping important concepts inside the chapter properly. You can also download our app Studyrankers for offline NCERT Solutions and learn efficiently.

NCERT Solutions is a better guide to increase concentration among students. In Exercise 1.3, you have to find products of given numbers.

NCERT Solutions for Class 7 Maths Ch 1 Integers Exercise 1.4

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NCERT Solutions for Class 7 Maths Ch 1 Integers Exercise 1.4

Class 7 Maths NCERT Solutions of Chapter 1 Integers Exercise 1.4 is useful in grasping key concepts inside the chapter properly. These NCERT Solutions are prepared by Studyrankers experts which are detailed and accurate. You can also download our app Studyrankers for offline NCERT Solutions and learn efficiently.

NCERT Solutions for Class 7 Maths provided on Studyrankers will inculcate correct learning habits among students. In Exercise 1.4, you have to evaluate the given expressions.

NCERT Solutions for Class 7 Maths Ch 2 Fractions and Decimals Exercise 2.1

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NCERT Solutions for Class 7 Maths Ch 2 Fractions and Decimals Exercise 2.1

Chapter 2 Fractions and Decimals Exercise 2.1 Class 7 Maths NCERT Solutions is very helpful in grasping key concepts inside the chapter properly. Given NCERT Solutions are useful in clearing doubts and completing your homework on time and clear your doubts. Through the help of these NCERT Solutions, one can clear their doubts and inculcate correct learning habits.

In Exercise 2.1, you have to arrange the given numbers in descending order and various word problems.