case study for quadratic equation class 10

Case Study Questions Class 10 Maths Quadratic Equations

Case study questions class 10 maths chapter 4 quadratic equations.

CBSE Class 10 Case Study Questions Maths Quadratic Equations. Term 2 Important Case Study Questions for Class 10 Board Exam Students. Here we have arranged some Important Case Base Questions for students who are searching for Paragraph Based Questions Quadratic Equations.

CBSE Case Study Questions Class 10 Maths Quadratic Equations

Raj and Ajay are very close friends. Both the families decide to go to Ranikhet by their own cars. Raj’s car travels at a speed of x km/h while Ajay’s car travels 5 km/h faster than Raj’s car. Raj took 4 hours more than Ajay to complete the journey of 400 km.

1.) What will be the distance covered by Ajay’s car in two hours?

3.) What is the speed of Raj’s car?

a) 20 km/hour

Answer – a) 20 km/hour

Q.2) Nidhi and Riya are very close friends. Nidhi’s parents have a Maruti Alto. Riya ‘s parents have a Toyota. Both the families decided to go for a picnic to Somnath Temple in Gujarat by their own car. Nidhi’s car travels x km/h, while Riya’s car travels 5km/h more than Nidhi’s car. Nidhi’s car took 4 hours more than Riya’s car in covering 400 km.

(ii) Write the quadratic equation describe the speed of Nidhi’s car. What is the speed of Nidhi’s car?

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CBSE Case Study Questions for Class 10 Maths Quadratic Equation Free PDF

Mere Bacchon, you must practice the CBSE Case Study Questions Class 10 Maths Quadratic Equation  in order to fully complete your preparation . They are very very important from exam point of view. These tricky Case Study Based Questions can act as a villain in your heroic exams!

I have made sure the questions (along with the solutions) prepare you fully for the upcoming exams. To download the latest CBSE Case Study Questions , just click ‘ Download PDF ’.

CBSE Case Study Questions for Class 10 Maths Quadratic Equation PDF

Checkout our case study questions for other chapters.

• Chapter 2: Polynomials Case Study Questions
• Chapter 3: Pair of Linear Equations in Two Variables Case Study Questions
• Chapter 5: Arithmetic Progressions Case Study Questions
• Chapter 6: Triangles Case Study Questions

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CBSE Class 10 Maths Case Study Questions PDF

Download Case Study Questions for Class 10 Mathematics to prepare for the upcoming CBSE Class 10 Final Exam. These Case Study and Passage Based questions are published by the experts of CBSE Experts for the students of CBSE Class 10 so that they can score 100% on Boards.

CBSE Class 10 Mathematics Exam 2024  will have a set of questions based on case studies in the form of MCQs. The CBSE Class 10 Mathematics Question Bank on Case Studies, provided in this article, can be very helpful to understand the new format of questions. Share this link with your friends.

Table of Contents

Chapterwise Case Study Questions for Class 10 Mathematics

Inboard exams, students will find the questions based on assertion and reasoning. Also, there will be a few questions based on case studies. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

The above  Case studies for Class 10 Maths will help you to boost your scores as Case Study questions have been coming in your examinations. These CBSE Class 10 Mathematics Case Studies have been developed by experienced teachers of cbseexpert.com for the benefit of Class 10 students.

• Class 10th Science Case Study Questions
• Assertion and Reason Questions of Class 10th Science
• Assertion and Reason Questions of Class 10th Social Science

Class 10 Maths Syllabus 2024

Chapter-1  real numbers.

Starting with an introduction to real numbers, properties of real numbers, Euclid’s division lemma, fundamentals of arithmetic, Euclid’s division algorithm, revisiting irrational numbers, revisiting rational numbers and their decimal expansions followed by a bunch of problems for a thorough and better understanding.

Chapter-2  Polynomials

This chapter is quite important and marks securing topics in the syllabus. As this chapter is repeated almost every year, students find this a very easy and simple subject to understand. Topics like the geometrical meaning of the zeroes of a polynomial, the relationship between zeroes and coefficients of a polynomial, division algorithm for polynomials followed with exercises and solved examples for thorough understanding.

Chapter-3  Pair of Linear Equations in Two Variables

This chapter is very intriguing and the topics covered here are explained very clearly and perfectly using examples and exercises for each topic. Starting with the introduction, pair of linear equations in two variables, graphical method of solution of a pair of linear equations, algebraic methods of solving a pair of linear equations, substitution method, elimination method, cross-multiplication method, equations reducible to a pair of linear equations in two variables, etc are a few topics that are discussed in this chapter.

Chapter-4  Quadratic Equations

The Quadratic Equations chapter is a very important and high priority subject in terms of examination, and securing as well as the problems are very simple and easy. Problems like finding the value of X from a given equation, comparing and solving two equations to find X, Y values, proving the given equation is quadratic or not by knowing the highest power, from the given statement deriving the required quadratic equation, etc are few topics covered in this chapter and also an ample set of problems are provided for better practice purposes.

Chapter-5  Arithmetic Progressions

This chapter is another interesting and simpler topic where the problems here are mostly based on a single formula and the rest are derivations of the original one. Beginning with a basic brief introduction, definitions of arithmetic progressions, nth term of an AP, the sum of first n terms of an AP are a few important and priority topics covered under this chapter. Apart from that, there are many problems and exercises followed with each topic for good understanding.

Chapter-6  Triangles

This chapter Triangle is an interesting and easy chapter and students often like this very much and a securing unit as well. Here beginning with the introduction to triangles followed by other topics like similar figures, the similarity of triangles, criteria for similarity of triangles, areas of similar triangles, Pythagoras theorem, along with a page summary for revision purposes are discussed in this chapter with examples and exercises for practice purposes.

Chapter-7  Coordinate Geometry

Here starting with a general introduction, distance formula, section formula, area of the triangle are a few topics covered in this chapter followed with examples and exercises for better and thorough practice purposes.

Chapter-8  Introduction to Trigonometry

As trigonometry is a very important and vast subject, this topic is divided into two parts where one chapter is Introduction to Trigonometry and another part is Applications of Trigonometry. This Introduction to Trigonometry chapter is started with a general introduction, trigonometric ratios, trigonometric ratios of some specific angles, trigonometric ratios of complementary angles, trigonometric identities, etc are a few important topics covered in this chapter.

Chapter-9  Applications of Trigonometry

This chapter is the continuation of the previous chapter, where the various modeled applications are discussed here with examples and exercises for better understanding. Topics like heights and distances are covered here and at the end, a summary is provided with all the important and frequently used formulas used in this chapter for solving the problems.

Chapter-10  Circle

Beginning with the introduction to circles, tangent to a circle, several tangents from a point on a circle are some of the important topics covered in this chapter. This chapter being practical, there are an ample number of problems and solved examples for better understanding and practice purposes.

Chapter-11  Constructions

This chapter has more practical problems than theory-based definitions. Beginning with a general introduction to constructions, tools used, etc, the topics like division of a line segment, construction of tangents to a circle, and followed with few solved examples that help in solving the exercises provided after each topic.

Chapter-12  Areas related to Circles

This chapter problem is exclusively formula based wherein topics like perimeter and area of a circle- A Review, areas of sector and segment of a circle, areas of combinations of plane figures, and a page summary is provided just as a revision of the topics and formulas covered in the entire chapter and also there are many exercises and solved examples for practice purposes.

Chapter-13  Surface Areas and Volumes

Starting with the introduction, the surface area of a combination of solids, the volume of a combination of solids, conversion of solid from one shape to another, frustum of a cone, etc are to name a few topics explained in detail provided with a set of examples for a better comprehension of the concepts.

Chapter-14  Statistics

In this chapter starting with an introduction, topics like mean of grouped data, mode of grouped data, a median of grouped, graphical representation of cumulative frequency distribution are explained in detail with exercises for practice purposes. This chapter being a simple and easy subject, securing the marks is not difficult for students.

Chapter-15  Probability

Probability is another simple and important chapter in examination point of view and as seeking knowledge purposes as well. Beginning with an introduction to probability, an important topic called A theoretical approach is explained here. Since this chapter is one of the smallest in the syllabus and problems are also quite easy, students often like this chapter

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Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations

Previous Year Questions 2024

Q1: In flight of 2800 km, an aircraft was slowed down due to bad weather. Its average speed is reduced by100km/h and by doing so, the time of flight is increased by 30 minutes. Find the original duration of the flight.     (2024)

Q9: If the sum of the roots of the quadratic equation ky 2 – 11y + (k – 23) = 0 is 13/21 more than the product of the roots, then find the value of k.    (2022)

Q21: Find the value of k for which x = 2 is a solution of the equation kx 2 + 2x - 3 = 0.     (2019)

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Class 10th Maths - Quadratic Equations Case Study Questions and Answers 2022 - 2023

Class 10th Maths - Quadratic Equations Case Study Questions and Answers 2022 - 2023 Study Materials Sep-09 , 2022

QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 10th Maths Subject - Quadratic Equations, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.

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Quadratic equations case study questions with answer key.

Final Semester - June 2015

A quadratic equation can be defined as an equation of degree 2. This means that the highest exponent of the polynomial in it is 2. The standard form of a quadratic equation is ax 2 + bx + c = 0, where a, b, and c are real numbers and  \(a \neq 0\)   Every quadratic equation has two roots depending on the nature of its discriminant, D = b2 - 4ac.Based on the above information, answer the following questions. (i) Which of the following quadratic equation have no real roots?

(ii) Which of the following quadratic equation have rational roots?

(iii) Which of the following quadratic equation have irrational roots?

(iv) Which of the following quadratic equations have equal roots?

(v) Which of the following quadratic equations has two distinct real roots?

In our daily life we use quadratic formula as for calculating areas, determining a product's profit or formulating the speed of an object and many more. Based on the above information, answer the following questions. (i) If the roots of the quadratic equation are 2, -3, then its equation is

(ii) If one root of the quadratic equation 2x 2 + kx + 1 = 0 is -1/2, then k =

(iii) Which of the following quadratic equations, has equal and opposite roots?

(iv) Which of the following quadratic equations can be represented as (x - 2) 2 + 19 = 0?

(v) If one root of a qua drraattiic equation is  \(\frac{1+\sqrt{5}}{7}\) , then I.ts other root is

Quadratic equations started around 3000 B.C. with the Babylonians. They were one of the world's first civilisation, and came up with some great ideas like agriculture, irrigation and writing. There were many reasons why Babylonians needed to solve quadratic equations. For example to know what amount of crop you can grow on the square field; Based on the above information, represent the following questions in the form of quadratic equation. (i) The sum of squares of two consecutive integers is 650.

(ii) The sum of two numbers is 15 and the sum of their reciprocals is 3/10.

(iii) Two numbers differ by 3 and their product is 504.

(iv) A natural number whose square diminished by 84 is thrice of 8 more of given number.

(v) A natural number when increased by 12, equals 160 times its reciprocal.

Amit is preparing for his upcoming semester exam. For this, he has to practice the chapter of Quadratic Equations. So he started with factorization method. Let two linear factors of  \(a x^{2}+b x+c \text { be }(p x+q) \text { and }(r x+s)\) \(\therefore a x^{2}+b x+c=(p x+q)(r x+s)=p r x^{2}+(p s+q r) x+q s .\) Now, factorize each of the following quadratic equations and find the roots. (i) 6x 2 + x - 2 = 0

(ii) 2x 2 -+ x - 300 = 0

(iii) x 2 -  8x + 16 = 0

(iv) 6x 2 -  13x + 5 = 0

(v) 100x 2 - 20x + 1 = 0

If p(x) is a quadratic polynomial i.e., p(x) = ax 2 - + bx + c, \(a \neq 0\) , then p(x) = 0 is called a quadratic equation. Now, answer the following questions. (i) Which of the following is correct about the quadratic equation ax 2 - + bx + c = 0 ?

(ii) The degree of a quadratic equation is

(iii) Which of the following is a quadratic equation?

(iv) Which of the following is incorrect about the quadratic equation ax 2 - + bx + c = 0 ?

(v) Which of the following is not a method of finding solutions of the given quadratic equation?

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Chapter 4 Class 10 Quadratic Equations

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Get NCERT Solutions for all exercise questions and examples of Chapter 4 Class 10 Quadratic Equations free at Teachoo. Answers to each and every question is provided video solutions. 

In this chapter, we will learn

• What is a Quadratic Equation
• What is the Standard Form of a Quadratic Equation
• Solution of a Quadratic Equation by Factorisation ( Splitting the Middle Term method)
• Solving a Quadratic Equation by Completing the Square
• Solving a Quadratic Equation using D Formula (x = -b ± √b 2 - 4ac / 2a)
• Checking if roots are real, equal or no real roots (By Checking the value of D = b 2 - 4ac)

This chapter is divided into two parts - Serial Order Wise, Concept Wise

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• Mathematics CBSE
• Quadratic Equations
• Nature of Roots

12. Case study: Nature of roots

Exercise condition:.

• 2 x 2 − 4 x + 390 = 0
• 2 x 2 − 4 x − 390 = 0
• 2 x 2 + 4 x − 390 = 0
• 2 x 2 + 4 x + 390 = 0
• \(ax^2 - bx + c = 0\)
• \(ax^2 + bx + c = 0\)
• \(ax^2 + bx = c\)
• \(ax^2 + ax + c = 0\)
• \(b^2 - 4ac = 0\)
• \(b^2 - 4ac \le 0\)
• \(b^2 - 4ac < 0\)
• \(b^2 - 4ac > 0\)

• Math Article
• Quadratic Equation For Class 10

Quadratic Equation Class 10 Notes Chapter 4

Cbse class 10 maths quadratic equation notes:- download pdf here.

Get the complete concepts covered in quadratic equations for Class 10 Maths here. These quadratic equations notes help the students to recall the important definitions, formulas and tricks to solve the problems in the CBSE Board Exams 2023-24. In this article, you will learn the concept of quadratic equations, standard form, nature of roots, and methods for finding the solution for the given quadratic equations with more examples.

Introduction to Quadratic Equations

Quadratic polynomial.

A polynomial of the form a x 2 + b x + c , where a, b and c are real numbers and a ≠ 0 is called a quadratic polynomial.

For more information on Quadratic Polynomials, watch the below video

Quadratic Equation

When we equate a quadratic polynomial to a constant, we get a quadratic equation.

Any equation of the form p(x) = k, where p(x) is a polynomial of degree 2, and c is a constant, is a quadratic equation.

The Standard Form of a Quadratic Equation

The standard form of a quadratic equation is ax 2 +bx+c=0, where a,b and c are real numbers and a≠0. ‘a’ is the coefficient of x 2 . It is called the quadratic coefficient. ‘b’ is the coefficient of x. It is called the linear coefficient. ‘c’ is the constant term.

To know more about Quadratic Equations, visit here .

Solving Quadratic Equations by Factorisation

Roots of a quadratic equation.

The values of x for which a quadratic equation is satisfied are called the roots of the quadratic equation.

If α is a root of the quadratic equation ax 2 +bx+c=0, then aα 2 +bα+c=0.

A quadratic equation can have two distinct real roots, two equal roots or real roots may not exist.

Graphically, the roots of a quadratic equation are the points where the graph of the quadratic polynomial cuts the x-axis.

Consider the graph of a quadratic equation x 2 −4=0

In the above figure, -2 and 2 are the roots of the quadratic equation x 2 −4=0 Note:

• If the graph of the quadratic polynomial cuts the x-axis at two distinct points, then it has real and distinct roots.
• If the graph of the quadratic polynomial touches the x-axis, then it has real and equal roots.
• If the graph of the quadratic polynomial does not cut or touch the x-axis then it does not have any real roots.

Solving a Quadratic Equation by Factorization Method

Consider a quadratic equation 2x 2 −5x+3=0

⇒2x 2 −2x−3x+3=0

This step is splitting the middle term.

We split the middle term by finding two numbers (-2 and -3) such that their sum is equal to the coefficient of x and their product is equal to the product of the coefficient of x 2 , and the constant.

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

And (-2) × (-3) = 6

2x 2 −2x−3x+3=0

2x(x−1)−3(x−1)=0

(x−1)(2x−3)=0

In this step, we have expressed the quadratic polynomial as a product of its factors.

Thus, x = 1 and x =3/2 are the roots of the given quadratic equation.

This method of solving a quadratic equation is called the factorisation method.

For more information on Solving a Quadratic Equation by Factorization Method, watch the below video

To know more about Solving Quadratic Equation by Factorisation, visit here .

Solving a Quadratic Equation by Completion of Squares Method

In the method of completing the squares, the quadratic equation is expressed in the form (x±k) 2 =p 2 .

Consider the quadratic equation 2x 2 −8x=10 (i) Express the quadratic equation in standard form. 2x 2 −8x−10=0

(ii) Divide the equation by the coefficient of x 2 to make the coefficient of x 2 equal to 1. x 2 −4x−5=0

(iii) Add the square of half of the coefficient of x to both sides of the equation to get an expression of the form x 2 ±2kx+k 2 . (x 2 −4x+4)−5=0+4

(iv) Isolate the above expression, (x±k) 2 on the LHS to obtain an equation of the form (x±k) 2 =p 2 (x−2) 2 =9

(v) Take the positive and negative square roots. x−2=±3

x=−1 or x=5

To know more about Solving Quadratic Equations by Completing the Square, visit here .

Solving Quadratic Equation Using Quadratic Formula

Quadratic formula.

Quadratic Formula is used to directly obtain the roots of a quadratic equation from the standard form of the equation.

For the quadratic equation ax 2 +bx+c=0,

x= [-b± √(b 2 -4ac)]/2a

By substituting the values of a,b and c, we can directly get the roots of the equation.

Example: If x 2 – 5x + 6 = 0 is the quadratic equation, find the roots.

Solution: Given, x 2 – 5x + 6 = 0 is the quadratic equation.

On comparing with the standard quadratic equation, we have;

ax 2 + bx + c = 0

a = 1, b = -5 and c = 6

b 2 – 4ac = (-5) 2 – 4 × 1 × 6 = 25 – 24 = 1 > 0 Hence, the roots are real. Using quadratic formula,

x = [-b ± √(b 2 – 4ac)]/ 2a

= [-(-5) ± √1]/ 2(1)

= [5 ± 1]/ 2

i.e. x = (5 + 1)/2 and x = (5 – 1)/2

x = 6/2, x = 4/2

Therefore, the roots of the quadratic equation are 3 and 2.

To know more about Quadratic Formula, visit here .

Discriminant

For a quadratic equation of the form ax 2 +bx+c=0, the expression b 2 −4ac is called the discriminant (denoted by D) of the quadratic equation.

The discriminant determines the nature of the roots of the quadratic equation based on the coefficients of the quadratic equation.

For more information on Discriminant, watch the below video

To know more about Discriminant Formula, visit here .

Nature of Roots

Based on the value of the discriminant, D=b 2 −4ac, the roots of a quadratic equation, ax 2 + bx + c = 0, can be of three types.

Case 1: If D>0 , the equation has two distinct real roots .

Case 2: If D=0 , the equation has two equal real roots .

Case 3: If D<0 , the equation has no real roots .

Solving using Quadratic Formula when D>0

Solve 2x 2 −7x+3=0 using the quadratic formula.

(i) Identify the coefficients of the quadratic equation. a = 2,b = -7,c = 3

(ii) Calculate the discriminant, b 2 −4ac

D=(−7) 2 −4×2×3= 49-24 = 25

D> 0, therefore, the roots are distinct.

(iii) Substitute the coefficients in the quadratic formula to find the roots

x= [-(-7)± √((-7) 2 -4(2)(3))]/2(2)

x=3 and x= 1/2 are the roots.

Solving Quadratic Equation when D=0

Let us take an example of quadratic equation 3x 2 – 2x + 1/3 = 0.

Here, a = 3, b = -2 and c = 1/3

Determinant, D = b 2 – 4ac = (-2) 2 – 4 (3)(1/3) = 4 – 4 = 0

Thus, the given equation has equal roots.

Hence the roots are -b/2a and -b/2a, i.e., 1/3 and 1/3.

Solving Quadratic Equation when D < 0

Suppose the quadratic equation is 4x 2 + 3x + 5 = 0

Comparing with the standard form of quadratic equation, ax 2 + bx + c = 0,

a = 4, b = 3, c = 5

By the formula of determinant, we know;

Determinant (D) = b 2 – 4ac

= (3) 2 – 4(4)(5)

= -71 2 – 4ac)]/ 2a

= [-3 ± √(-71)]/ 2(4)

= [-3 ± √(i 2 71)]/ 8

= (-3 ± i√71)/8

Thus, the non-real roots of the equation are x = (-3 + i√71)/8 and x (-3 – i√71)/8.

For more information on Nature Of Roots, watch the below videos

To know more about the Nature of Roots, visit here .

Graphical Representation of a Quadratic Equation

The graph of a quadratic polynomial is a parabola. The roots of a quadratic equation are the points where the parabola cuts the x-axis i.e. the points where the value of the quadratic polynomial is zero.

Now, the graph of x 2 +5x+6=0 is:

In the above figure, -2 and -3 are the roots of the quadratic equation x 2 +5x+6=0.

For a quadratic polynomial ax 2 +bx+c,

If a>0, the parabola opens upwards. If a 2 −4ac

If D>0 , the parabola cuts the x-axis at exactly two distinct points. The roots are distinct. This case is shown in the above figure in a, where the quadratic polynomial cuts the x-axis at two distinct points.

If D=0 , the parabola just touches the x-axis at one point and the rest of the parabola lies above or below the x-axis. In this case, the roots are equal. This case is shown in the above figure in b, where the quadratic polynomial touches the x-axis at only one point .

If D<0 , the parabola lies entirely above or below the x-axis and there is no point of contact with the x-axis. In this case, there are no real roots. This case is shown in the above figure in c, where the quadratic polynomial neither cuts nor touches the x-axis.

Formation of a quadratic equation from its roots

To find out the standard form of a quadratic equation when the roots are given:

Let α and β be the roots of the quadratic equation ax 2 +bx+c=0. Then,

(x−α)(x−β)=0

On expanding, we get,

x 2 −(α+β)x+αβ=0, which is the standard form of the quadratic equation.

Here, a=1,b=−(α+β) and c=αβ.

Example: Form the quadratic equation if the roots are −3 and 4.

Solution: Given -3 and 4 are the roots of the equation.

Sum of roots = -3 + 4 = 1

Product of the roots = (-3).(4) = -12

As we know, the standard form of a quadratic equation is:

x 2 − (sum of roots)x + (product of roots) = 0

Therefore, by putting the values, we get

x 2 – x – 12 = 0

Which is the required quadratic equation.

Sum and Product of Roots of a Quadratic Equation

Sum of roots = α + β =-b/a

Product of roots = αβ = c/a

Example: Given, x 2 − 5x + 8 = 0 is the quadratic equation. Find the sum and product of its roots.

Solution: x 2 − 5x + 8 = 0 is the quadratic equation given in the form of ax 2 + bx + c = 0. Hence, a = 1 b = -5 c = 8

Sum of roots = -b/a = 5

Product of roots = c/a = 8

To know more about Sum and Product of Roots of a Quadratic equation, visit here .

For more information on Sum and Product of Roots of a Quadratic equation, watch the below video

Related Links:

• NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations
• NCERT Exemplar Class 10 Maths Solutions for Chapter 4 – Quadratic Equations
• RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations
• Class 10 Maths Chapter 4 Quadratic Equations MCQs
• Important Questions Class 10 Maths Chapter 4 Quadratic Equations

Practice Questions on Quadratic Equations Class 10

1. Check whether the following are quadratic equations: (i) (x – 2) 2 + 1 = 2x – 3 (ii) x(x + 1) + 8 = (x + 2) (x – 2) (iii) x (2x + 3) = x 2 + 1 (iv) (x + 2) 3 = x 3 – 4 2. Find two numbers whose sum is 27 and product is 182. 3. Find the roots of 4x 2 + 3x + 5 = 0 by the method of completing the square. 4. Find the roots of the quadratic equation 3x 2 – 5x + 2 = 0, if they exist, using the quadratic formula. 5. Find the values of k for which the quadratic equation kx(x – 2) + 6 = 0 has two equal roots.

Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin!

Select the correct answer and click on the “Finish” button Check your score and answers at the end of the quiz

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NCERT Solutions for Class 10 Maths Chapter 4 - Quadratic Equations

• NCERT Solutions
• Chapter 4 Quadratic Equations

Complete Resource of NCERT Class 10 Maths Chapter 4 Quadratic Equations - Free PDF Download

NCERT Solutions for Class 10 Chapter 4 Quadratic Equation , covers a crucial aspect of algebra that every student must grasp. This chapter introduces quadratic equations, detailing methods to solve them, including factoring, using the quadratic formula, and completing the square. Understanding these concepts is key, as they are foundational for higher mathematics and problem solving in science and engineering. Focus on mastering the techniques for finding the roots of quadratic equations and recognizing their practical applications. Clear explanations and step-by-step solutions in Vedantu’s materials help understand complex concepts, making them accessible and understandable.

Glance of NCERT Solutions of Maths Chapter 4 Quadratic Equations for Class 10 | Vedantu

Chapter 4 of Class 10 Maths deals with quadratic equations, which are equations  of the form  ax^2 + bx + c = 0, where a ≠ 0.

Learn about standard forms, where a, b, and c are real numbers.

The chapter focuses on finding the roots/solutions of these equations, which are the values of x that satisfy the equation.

There are different methods for solving quadratics, such as Factorization and by using Quadratic Formula

A key concept is the discriminant (b² - 4ac). It helps determine the nature of the roots:

Distinct Real Roots (D > 0): When the discriminant is positive, there are two distinct real solutions for x.

Equal Real Roots (D = 0): A positive discriminant of zero indicates two equal real roots.

No Real Roots (D < 0): A negative discriminant means there are no real number solutions, but there might be complex solutions.

The chapter also covers forming quadratic equations from word problems and applications of quadratic equations in real-life scenarios.

This article contains chapter notes important questions and Exercises link for Chapter 4 - Quadratic Equations, which you can download as PDFs.

There are four exercises and one miscellaneous exercise (24 fully solved questions) in class 10th maths chapter 4 Quadratic Equations.

Access Exercise Wise NCERT Solutions for Chapter 4 Maths Class 10

Related Chapters

Exercises under NCERT Solutions for Maths Chapter 4 Class 10 Quadratic Equations

Exercise 4.1:.

This exercise covers the introduction to quadratic equations and the standard form of a quadratic equation. It also includes methods for solving quadratic equations by factorisation. In this exercise, students will learn how to identify a quadratic equation, how to convert a quadratic equation into standard form, and how to factorise quadratic equations using different methods. The exercise includes a set of questions that range from easy to difficult, allowing students to gradually build their understanding of the concepts.

Exercise 4.2:

This exercise covers more advanced methods for solving quadratic equations, such as completing the square and using the quadratic formula. It includes the derivation of the quadratic formula and shows how to apply it to solve quadratic equations. In this exercise, students will learn how to complete the square of a quadratic equation to convert it into standard form, and how to use the quadratic formula to solve quadratic equations. The exercise includes questions that require students to use both methods to solve quadratic equations.

Exercise 4.3:

This exercise covers real-life applications of quadratic equations and includes word problems that require students to apply their knowledge of quadratic equations to solve practical problems. The exercise includes problems related to the trajectory of a projectile, finding the distance between two ships, and the dimensions of a garden. Students will learn how to formulate and solve quadratic equations to solve real-life problems. The exercise includes a set of word problems that gradually increase in difficulty, allowing students to develop their problem-solving skills.

Access NCERT Solutions for Class - 10 Maths Chapter 4 – Quadratic Equations

Exercise 4.1.

1. Check whether the following are quadratic equations:  

i. ${{\left( \text{x+1} \right)}^{\text{2}}}\text{=2}\left( \text{x-3} \right)$

Ans : ${{\left( \text{x+1} \right)}^{\text{2}}}\text{=2}\left( \text{x-3} \right)$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+2x+1=2x-6}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+7=0}$

Since, it is in the form of $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=0}$.

Therefore, the given equation is a quadratic equation.

ii. ${{\text{x}}^{\text{2}}}\text{-2x=}\left( \text{-2} \right)\left( \text{3-x} \right)$

Ans : ${{\text{x}}^{\text{2}}}\text{-2x=}\left( \text{-2} \right)\left( \text{3-x} \right)$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-2x=-6+2x}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-4x+6=0}$

iii. $\left( \text{x-2} \right)\left( \text{x+1} \right)\text{=}\left( \text{x-1} \right)\left( \text{x+3} \right)$

Ans : $\left( \text{x-2} \right)\left( \text{x+1} \right)\text{=}\left( \text{x-1} \right)\left( \text{x+3} \right)$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-x-2=}{{\text{x}}^{\text{2}}}\text{+2x-3}$

$\Rightarrow \text{3x-1=0}$

Since, it is not in the form of $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=0}$.

Therefore, the given equation is not a quadratic equation.

iv. $\left( \text{x-3} \right)\left( \text{2x+1} \right)\text{=x}\left( \text{x+5} \right)$

Ans : $\left( \text{x-3} \right)\left( \text{2x+1} \right)\text{=x}\left( \text{x+5} \right)$

$\Rightarrow \text{2}{{\text{x}}^{\text{2}}}\text{-5x-3=}{{\text{x}}^{\text{2}}}\text{+5x}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-10x-3=0}$

v. $\left( \text{2x-1} \right)\left( \text{x-3} \right)\text{=}\left( \text{x+5} \right)\left( \text{x-1} \right)$

Ans : $\left( \text{2x-1} \right)\left( \text{x-3} \right)\text{=}\left( \text{x+5} \right)\left( \text{x-1} \right)$

$\Rightarrow \text{2}{{\text{x}}^{\text{2}}}\text{-7x+3=}{{\text{x}}^{\text{2}}}\text{+4x-5}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-11x+8=0}$

vi. ${{\text{x}}^{\text{2}}}\text{+3x+1=}{{\left( \text{x-2} \right)}^{\text{2}}}$

Ans : ${{\text{x}}^{\text{2}}}\text{+3x+1=}{{\left( \text{x-2} \right)}^{\text{2}}}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+3x+1=}{{\text{x}}^{\text{2}}}\text{+4-4x}$

$\Rightarrow \text{7x-3=0}$

vii. ${{\left( \text{x+2} \right)}^{\text{3}}}\text{=2x}\left( {{\text{x}}^{\text{2}}}\text{-1} \right)$

Ans : ${{\left( \text{x+2} \right)}^{\text{3}}}\text{=2x}\left( {{\text{x}}^{\text{2}}}\text{-1} \right)$

$\Rightarrow {{\text{x}}^{\text{3}}}\text{+8+6}{{\text{x}}^{\text{2}}}\text{+12x=2}{{\text{x}}^{\text{3}}}\text{-2x}$

$\Rightarrow {{\text{x}}^{\text{3}}}\text{-14x-6}{{\text{x}}^{\text{2}}}\text{-8=0}$

viii. ${{\text{x}}^{\text{3}}}\text{-4}{{\text{x}}^{\text{2}}}\text{-x+1=}{{\left( \text{x-2} \right)}^{\text{3}}}$

Ans : ${{\text{x}}^{\text{3}}}\text{-4}{{\text{x}}^{\text{2}}}\text{-x+1=}{{\left( \text{x-2} \right)}^{\text{3}}}$ 

$\Rightarrow {{\text{x}}^{\text{3}}}\text{-4}{{\text{x}}^{\text{2}}}\text{-x+1=}{{\text{x}}^{\text{3}}}\text{-8-6}{{\text{x}}^{\text{2}}}\text{+12x}$

$\Rightarrow \text{2}{{\text{x}}^{\text{2}}}\text{-13x+9=0}$

2. Represent the following situations in the form of quadratic equations.

i. The area of a rectangular plot is $\text{528 }{{\text{m}}^{\text{2}}}$. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.

Ans : Let the breath of the plot be $\text{x m}$.

Thus, length would be-

$\text{Length=}\left( \text{2x+1} \right)\text{m}$

Hence, Area of rectangle $=$$\text{Length }\!\!\times\!\!\text{ breadth}$

So, $\text{528=x}\left( \text{2x+1} \right)$

$\Rightarrow \text{2}{{\text{x}}^{\text{2}}}\text{+x-528=0}$

ii. The product of two consecutive positive integers is $\text{306}$. We need to find the integers.

Ans : Let the consecutive integers be $\text{x}$ and $\text{x+1}$.

Thus, according to question-

$\text{x}\left( \text{x+1} \right)\text{=306}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+x-306=0}$

iii. Rohan’s mother is $\text{26}$ years older than him. The product of their ages (in years) $\text{3}$ years from now will be $\text{360}$. We would like to find Rohan’s present age.

Ans : Let Rohan’s age be $\text{x}$.

Hence, his mother’s age is $\text{x+26}$ .

Now, after $\text{3 years}$.

Rohan’s age will be $\text{x+3}$.

His mother’s age will be $\text{x+29}$ .

So, according to question-

$\left( \text{x+3} \right)\left( \text{x+29} \right)\text{=360}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+3x+29x+87=360}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+32x-273=0}$

iv. A train travels a distance of $\text{480 km}$ at a uniform speed. If the speed had been $\text{8km/h}$ less, then it would have taken $\text{3}$ hours more to cover the same distance. We need to find the speed of the train.

Ans : Let the speed of train be $\text{x km/h}$.

Thus, time taken to travel $\text{482 km}$ is $\dfrac{\text{480}}{\text{x}}\text{hrs}$.

Now, let the speed of train $\text{=}\left( \text{x-8} \right)\text{km/h}$.

Therefore, time taken to travel $\text{480 km}$ is $\left( \dfrac{\text{480}}{\text{x}}+3 \right)\text{hrs}$.

Hence, $\text{speed }\!\!\times\!\!\text{ time=distance}$

i.e $\left( \text{x-8} \right)\left( \dfrac{\text{480}}{\text{x}}\text{+3} \right)\text{=480}$

$\Rightarrow \text{480+3x-}\dfrac{\text{3840}}{\text{x}}\text{-24=480}$

$\Rightarrow \text{3x-}\dfrac{\text{3840}}{\text{x}}\text{=24}$

$\Rightarrow \text{3}{{\text{x}}^{\text{2}}}\text{-24x-3840=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-8x-1280=0}$

Exercise 4.2

1. Find the roots of the following quadratic equations by factorisation:

i. ${{\text{x}}^{\text{2}}}\text{-3x-10=0}$

Ans : ${{\text{x}}^{\text{2}}}\text{-3x-10=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-5x+2x-10}$

$\Rightarrow \text{x}\left( \text{x-5} \right)\text{+2}\left( \text{x-5} \right)$

$\Rightarrow \left( \text{x-5} \right)\left( \text{x+2} \right)$

Therefore, roots of this equation are –

$\text{x-5=0}$ or $\text{x+2=0}$

i.e $\text{x=5}$ or $\text{x=-2}$

ii. $\text{2}{{\text{x}}^{\text{2}}}\text{+x-6=0}$

Ans : $\text{2}{{\text{x}}^{\text{2}}}\text{+x-6=0}$

$\Rightarrow 2{{\text{x}}^{\text{2}}}\text{+4x-3x-6}$

$\Rightarrow 2\text{x}\left( \text{x+2} \right)-3\left( \text{x+2} \right)$

$\Rightarrow \left( \text{x+2} \right)\left( \text{2x-3} \right)$

$\text{x+2=0}$ or $\text{2x-3=0}$

i.e $\text{x=-2}$ or $\text{x=}\dfrac{3}{2}$

iii. $\sqrt{\text{2}}{{\text{x}}^{\text{2}}}\text{+7x+5}\sqrt{\text{2}}\text{=0}$

Ans : $\sqrt{\text{2}}{{\text{x}}^{\text{2}}}\text{+7x+5}\sqrt{\text{2}}\text{=0}$

$\Rightarrow \sqrt{\text{2}}{{\text{x}}^{\text{2}}}\text{+5x+2x+5}\sqrt{\text{2}}$

$\Rightarrow \text{x}\left( \sqrt{\text{2}}\text{x+5} \right)+\sqrt{\text{2}}\left( \sqrt{\text{2}}\text{x+5} \right)$

$\Rightarrow \left( \sqrt{\text{2}}\text{x+5} \right)\left( \text{x+}\sqrt{\text{2}} \right)$

$\sqrt{\text{2}}\text{x+5=0}$ or $\text{x+}\sqrt{\text{2}}\text{=0}$

i.e $\text{x=}\dfrac{-5}{\sqrt{\text{2}}}$ or $\text{x=-}\sqrt{\text{2}}$

iv. $\text{2}{{\text{x}}^{\text{2}}}\text{-x+}\dfrac{\text{1}}{\text{8}}\text{=0}$

Ans : $\text{2}{{\text{x}}^{\text{2}}}\text{-x+}\dfrac{\text{1}}{\text{8}}\text{=0}$

\[\Rightarrow \dfrac{\text{1}}{\text{8}}\left( 16{{\text{x}}^{\text{2}}}-8x+1 \right)\]

\[\Rightarrow \dfrac{\text{1}}{\text{8}}\left( 4x\left( 4x-1 \right)-1\left( 4x-1 \right) \right)\]

$\Rightarrow \dfrac{\text{1}}{\text{8}} {{\left( \text{4x-1} \right)}^{2}}$

$\text{4x-1=0}$ or $\text{4x-1=0}$

i.e $\text{x=}\dfrac{1}{4}$ or $\text{x=}\dfrac{1}{4}$

v. $\text{100}{{\text{x}}^{\text{2}}}\text{-20x+1=0}$

Ans : $\text{100}{{\text{x}}^{\text{2}}}\text{-20x+1=0}$

$\Rightarrow 100{{\text{x}}^{\text{2}}}\text{-10x-10x+1}$

$\Rightarrow 10\text{x}\left( \text{10x-1} \right)-1\left( \text{10x-1} \right)$

\[\Rightarrow \left( \text{10x-1} \right)\left( \text{10x-1} \right)\]

\[\left( \text{10x-1} \right)=0\]or \[\left( \text{10x-1} \right)=0\]

i.e $\text{x=}\dfrac{1}{10}$ or $\text{x=}\dfrac{1}{10}$

2. Solve the problems given in Example 1

i. John and Jivanti together have $\text{45}$ marbles. Both of them lost $\text{5}$ marbles each, and the product of the number of marbles they now have is $\text{124}$. Find out how many marbles they had to start with.

Ans : Let the number of john’s marbles be $\text{x}$.

Thus, number of Jivanti’s marble be $\text{45-x}$.

According to question i.e, 

After losing $\text{5}$ marbles.

Number of john’s marbles be $\text{x-5}$

And number of Jivanti’s marble be $\text{40-x}$.

Therefore, $\left( \text{x-5} \right)\left( \text{40-x} \right)\text{=124}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-45x+324=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-36x-9x+324=0}$

$\Rightarrow \text{x}\left( \text{x-36} \right)\text{-9}\left( \text{x-36} \right)\text{=0}$

$\Rightarrow \left( \text{x-36} \right)\left( \text{x-9} \right)\text{=0}$

Case 1 - If $\text{x-36=0}$ i.e $\text{x=36}$

So, the number of john’s marbles be $\text{36}$.

Thus, number of Jivanti’s marble be $\text{9}$.

Case 2 - If $\text{x-9=0}$ i.e $\text{x=9}$

So, the number of john’s marbles be $9$.

Thus, number of Jivanti’s marble be $36$.

ii. A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be $\text{55}$ minus the number of toys produced in a day. On a particular day, the total cost of production was Rs $\text{750}$. Find out the number of toys produced on that day.

Ans: Let the number of toys produced be $\text{x}$.

Therefore, Cost of production of each toy be $\text{Rs}\left( \text{55-x} \right)$.

Thus, $\left( \text{55-x} \right)\text{x=750}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-55x+750=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-25x-30x+750=0}$

$\Rightarrow \text{x}\left( \text{x-25} \right)-30\left( \text{x-25} \right)\text{=0}$

$\Rightarrow \left( \text{x-25} \right)\left( \text{x-30} \right)\text{=0}$

Case 1 - If $\text{x-25=0}$ i.e $\text{x=25}$

So, the number of toys be $25$.

Case 2 - If $\text{x-30=0}$ i.e $\text{x=30}$

So, the number of toys be $30$.

3. Find two numbers whose sum is $\text{27}$ and product is $\text{182}$ .

Ans: Let the first number be $\text{x}$ ,

Thus, the second number be $\text{27-x}$.

$\text{x}\left( \text{27-x} \right)\text{=182}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-27x+182=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-13x-14x+182=0}$

$\Rightarrow \text{x}\left( \text{x-13} \right)-14\left( \text{x-13} \right)\text{=0}$

$\Rightarrow \left( \text{x-13} \right)\left( \text{x-14} \right)\text{=0}$

Case 1 - If $\text{x-13=0}$ i.e $\text{x=13}$

So, the first number be $13$ ,

Thus, the second number be $\text{14}$.

Case 2 - If $\text{x-14=0}$ i.e $\text{x=14}$

So, the first number be $\text{14}$.

Thus, the second number be$13$.

4. Find two consecutive positive integers, sum of whose squares is $\text{365}$.

Ans: Let the consecutive positive integers be $\text{x}$ and $\text{x+1}$.

Thus, ${{\text{x}}^{\text{2}}}\text{+}{{\left( \text{x+1} \right)}^{\text{2}}}\text{=365}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+}{{\text{x}}^{\text{2}}}+1+2\text{x=365}$

$\Rightarrow 2{{\text{x}}^{\text{2}}}\text{+2x-364=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+x-182=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+14x-13x-182=0}$

$\Rightarrow \text{x}\left( \text{x+14} \right)-13\left( \text{x+14} \right)\text{=0}$

$\Rightarrow \left( \text{x+14} \right)\left( \text{x-13} \right)\text{=0}$

Case 1 - If $\text{x+14=0}$ i.e $\text{x=-14}$.

This case is rejected because number is positive.

Case 2 - If $\text{x-13=0}$ i.e $\text{x=13}$

So, the first number be $\text{13}$.

Thus, the second number be $14$.

Hence, the two consecutive positive integers are $\text{13}$ and $14$.

5. The altitude of a right triangle is $\text{7 cm}$ less than its base. If the hypotenuse is $\text{13 cm}$, find the other two sides.

Ans: Let the base of the right-angled triangle be $\text{x cm}$.

Its altitude be $\left( \text{x-7} \right)\text{cm}$.

Thus, by pythagores theorem-

$\text{bas}{{\text{e}}^{\text{2}}}\text{+altitud}{{\text{e}}^{\text{2}}}\text{=hypotenus}{{\text{e}}^{\text{2}}}$

\[\therefore {{\text{x}}^{\text{2}}}\text{+}{{\left( \text{x-7} \right)}^{\text{2}}}\text{=1}{{\text{3}}^{\text{2}}}\]

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+}{{\text{x}}^{\text{2}}}+49-14\text{x=169}$

$\Rightarrow 2{{\text{x}}^{\text{2}}}\text{-14x-120=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-7x-60=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+12x+5x-60=0}$

$\Rightarrow \text{x}\left( \text{x-12} \right)+5\left( \text{x-12} \right)\text{=0}$

$\Rightarrow \left( \text{x-12} \right)\left( \text{x+5} \right)\text{=0}$

Case 1 - If $\text{x-12=0}$ i.e $\text{x=12}$.

So, the base of the right-angled triangle be $\text{12 cm}$ and Its altitude be $\text{5cm}$

Case 2 - If $\text{x+5=0}$ i.e $\text{x=-5}$

This case is rejected because side is always positive.

Hence, the base of the right-angled triangle be $\text{12 cm}$ and Its altitude be $\text{5cm}$.

6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was $\text{3}$ more than twice the number of articles produced on that day. If the total cost of production on that day was Rs $\text{90}$, find the number of articles produced and the cost of each article.

Ans:  Let the number of articles produced be $\text{x}$.

Therefore, cost of production of each article be $\text{Rs}\left( \text{2x+3} \right)$.

Thus, $\text{x}\left( \text{2x+3} \right)\text{=90}$

$\Rightarrow 2{{\text{x}}^{\text{2}}}\text{+3x-90=0}$

$\Rightarrow 2{{\text{x}}^{\text{2}}}\text{+15x-12x-90=0}$

$\Rightarrow \text{x}\left( \text{2x+15} \right)-6\left( \text{2x+15} \right)\text{=0}$

$\Rightarrow \left( \text{2x+15} \right)\left( \text{x-6} \right)\text{=0}$

Case 1 - If $\text{2x-15=0}$ i.e $\text{x=}\dfrac{-15}{2}$.

This case is rejected because number of articles is always positive.

Case 2 - If $\text{x-6=0}$ i.e $\text{x=6}$

Hence, the number of articles produced be $6$.

Therefore, cost of production of each article be $\text{Rs15}$.

Exercise 4.3

1. Find the nature of the roots of the following quadratic equations.  If the real roots exist, find them-

i. $\text{2}{{\text{x}}^{\text{2}}}\text{-3x+5=0}$

Ans: For a quadratic equation $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=0}$.

Where Discriminant $\text{=}{{\text{b}}^{\text{2}}}\text{-4ac}$

Case 1- If ${{\text{b}}^{\text{2}}}\text{-4ac>0}$ then there will be two distinct real roots.

Case 2- If ${{\text{b}}^{\text{2}}}\text{-4ac=0}$ then there will be two equal real roots.

Case 3- If ${{\text{b}}^{\text{2}}}\text{-4ac<0}$ then there will be no real roots.

Thus, for $\text{2}{{\text{x}}^{\text{2}}}\text{-3x+5=0}$ .

On comparing this equation with $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=}0$.

So, $\text{a=2}$, $\text{b=-3}$, $\text{c=5}$.

Discriminant $\text{=}{{\left( \text{-3} \right)}^{\text{2}}}\text{-4}\left( \text{2} \right)\left( \text{5} \right)$

$\text{=9-40}$

$\text{=-31}$

Since, Discriminant: ${{\text{b}}^{\text{2}}}\text{-4ac < 0}$.

Therefore, there is no real root for the given equation.

ii. $\text{3}{{\text{x}}^{\text{2}}}\text{-4}\sqrt{\text{3}}\text{x+4=0}$

Case 1- If ${{\text{b}}^{\text{2}}}\text{-4ac > 0}$ then there will be two distinct real roots.

Case 3- If ${{\text{b}}^{\text{2}}}\text{-4ac < 0}$ then there will be no real roots.

Thus, for $\text{3}{{\text{x}}^{\text{2}}}\text{-4}\sqrt{\text{3}}\text{x+4=0}$ .

So, $\text{a=3}$, $\text{b=-4}\sqrt{\text{3}}$, $\text{c=4}$.

Discriminant $\text{=}{{\left( \text{-4}\sqrt{\text{3}} \right)}^{\text{2}}}\text{-4}\left( \text{3} \right)\left( \text{4} \right)$

$\text{=48-48}$

$\text{=0}$

Since, Discriminant: ${{\text{b}}^{\text{2}}}\text{-4ac=0}$.

Therefore, there is equal real root for the given equation and the roots are-

$\dfrac{\text{-b}}{\text{2a}}$ and $\dfrac{\text{-b}}{\text{2a}}$.

Hence, roots are-

$\dfrac{\text{-b}}{\text{2a}}\text{=}\dfrac{\text{-}\left( \text{-4}\sqrt{\text{3}} \right)}{\text{6}}$

$\text{=}\dfrac{\text{4}\sqrt{\text{3}}}{\text{6}}$

\[\text{=}\dfrac{\text{2}\sqrt{\text{3}}}{3}\]

Therefore, roots are \[\dfrac{\text{2}\sqrt{\text{3}}}{3}\] and \[\dfrac{\text{2}\sqrt{\text{3}}}{3}\].

iii. $\text{2}{{\text{x}}^{\text{2}}}\text{-6x+3=0}$

Thus, for $\text{2}{{\text{x}}^{\text{2}}}\text{-6x+3=0}$ .

So, $\text{a=2}$, $\text{b=-6}$, $\text{c=3}$.

Discriminant $\text{=}{{\left( \text{-6} \right)}^{\text{2}}}\text{-4}\left( \text{2} \right)\left( \text{3} \right)$

$\text{=36-24}$

$\text{=12}$

Since, Discriminant: ${{\text{b}}^{\text{2}}}\text{-4ac>0}$.

Therefore, distinct real roots exists for the given equation and the roots are-

$\text{x=}\dfrac{\text{-b }\!\!\pm\!\!\text{ }\sqrt{{{\text{b}}^{\text{2}}}\text{-4ac}}}{\text{2a}}$

$\text{x=}\dfrac{\text{-}\left( \text{-6} \right)\text{ }\!\!\pm\!\!\text{ }\sqrt{{{\left( \text{-6} \right)}^{\text{2}}}\text{-4}\left( \text{2} \right)\left( \text{3} \right)}}{\text{4}}$

$\text{=}\dfrac{\text{6 }\!\!\pm\!\!\text{ }\sqrt{\text{36-24}}}{\text{4}}$

$\text{=}\dfrac{\text{6 }\!\!\pm\!\!\text{ }\sqrt{\text{12}}}{\text{4}}$

$\text{=}\dfrac{\text{6 }\!\!\pm\!\!\text{ 2}\sqrt{\text{3}}}{\text{4}}$

$\text{=}\dfrac{\text{3 }\!\!\pm\!\!\text{ }\sqrt{\text{3}}}{\text{2}}$

Therefore, roots are $\dfrac{\text{3+}\sqrt{\text{3}}}{\text{2}}$ and $\dfrac{\text{3-}\sqrt{\text{3}}}{\text{2}}$.

2. Find the values of $\text{k}$ for each of the following quadratic equations, so  that they have two equal roots.

i. $\text{2}{{\text{x}}^{\text{2}}}\text{+kx+3=0}$

Ans: If a quadratic equation $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=0}$ has two equal roots, then its discriminant will be $\text{0}$ i.e., ${{\text{b}}^{\text{2}}}\text{-4ac=0}$

So, for $\text{2}{{\text{x}}^{\text{2}}}\text{+kx+3=0}$ .

So, $\text{a=2}$, $\text{b=k}$, $\text{c=3}$.

Discriminant $\text{=}{{\left( \text{k} \right)}^{\text{2}}}\text{-4}\left( \text{2} \right)\left( \text{3} \right)$

$\text{=}{{\text{k}}^{2}}-24$

For equal roots-

${{\text{b}}^{\text{2}}}\text{-4ac=0}$

$\therefore {{\text{k}}^{\text{2}}}\text{-24=0}$

$\Rightarrow {{\text{k}}^{\text{2}}}\text{=24}$

$\Rightarrow \text{k=}\sqrt{\text{24}}$

$\Rightarrow \text{k=}\pm \text{2}\sqrt{\text{6}}$

ii. $\text{kx}\left( \text{x-2} \right)\text{+6=0}$

So, for $\text{kx}\left( \text{x-2} \right)\text{+6=0}$

$\Rightarrow \text{k}{{\text{x}}^{\text{2}}}\text{-2kx+6=0}$

So, $\text{a=k}$, $\text{b=-2k}$, $\text{c=6}$.

Discriminant $\text{=}{{\left( \text{-2k} \right)}^{\text{2}}}\text{-4}\left( \text{k} \right)\left( \text{6} \right)$

$\text{=4}{{\text{k}}^{\text{2}}}\text{-24k}$

$\therefore \text{4}{{\text{k}}^{\text{2}}}\text{-24k=0}$

$\Rightarrow \text{4k}\left( \text{k-6} \right)\text{=0}$

$\Rightarrow \text{k=0 or k=6}$

But $\text{k}$ cannot be zero. Thus, this equation has two equal roots when $\text{k}$ should be $\text{6}$ .

3. Is it possible to design a rectangular mango grove whose length is  twice its breadth, and the area is $\text{800}{{\text{m}}^{\text{2}}}$ ? If so, find its length and breadth.

Ans: Let the breadth of mango grove be $\text{x}$.

So, length of mango grove will be $\text{2x}$.

Hence, Area of mango grove is $=\left( \text{2x} \right)\text{x}$

$\text{=2}{{\text{x}}^{\text{2}}}$.

So, $\text{2}{{\text{x}}^{\text{2}}}\text{=800}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{=400}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-400=0}$

So, $\text{a=1}$, $\text{b=0}$, $\text{c=400}$.

Discriminant $\text{=}{{\left( \text{0} \right)}^{\text{2}}}\text{-4}\left( \text{1} \right)\left( \text{-400} \right)$

$\text{=1600}$

Therefore, distinct real roots exist for the given equation and the roots are-

$\text{x=}\dfrac{\text{-}\left( 0 \right)\text{ }\!\!\pm\!\!\text{ }\sqrt{{{\left( 0 \right)}^{\text{2}}}\text{-4}\left( 1 \right)\left( -400 \right)}}{2}$

$\text{=}\dfrac{\pm \sqrt{\text{1600}}}{2}$

$\text{=}\dfrac{\text{ }\!\!\pm\!\!\text{ 40}}{2}$

$\text{=}\pm \text{20}$

Since, length cannot be negative.

Therefore, breadth of the mango grove is $\text{20m}$.

And length of the mango grove be $\text{2}\left( \text{20} \right)\text{m}$ i.e., $\text{40m}$.

4. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is $\text{20}$ years. Four years ago, the product of their ages in years was $\text{48}$.

Ans: Let the age of one friend be $\text{x years}$.

So, age of the other friend will be $\left( \text{20-x} \right)\text{years}$.

Thus, four years ago, the age of one friend be $\left( \text{x-4} \right)\text{years}$.

And age of the other friend will be $\left( \text{16-x} \right)\text{years}$.

Hence, according to question-

$\left( \text{x-4} \right)\left( \text{16-x} \right)\text{=48}$

$\Rightarrow \text{16x-64-}{{\text{x}}^{\text{2}}}\text{+4x=48}$

$\Rightarrow 20\text{x-112-}{{\text{x}}^{\text{2}}}\text{=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-20x+112-=0}$

So, $\text{a=1}$, $\text{b=-20}$, $\text{c=112}$.

Discriminant $\text{=}{{\left( \text{-20} \right)}^{\text{2}}}\text{-4}\left( \text{1} \right)\left( \text{112} \right)$

$\text{=400-448}$

$\text{=-48}$

Since, Discriminant: ${{\text{b}}^{\text{2}}}\text{-4ac <0}$.

Therefore, there is no real root for the given equation and hence, this situation is not possible.

5. Is it possible to design a rectangular park of perimeter $\text{80 m}$ and area $\text{400}{{\text{m}}^{\text{2}}}$? If so find its length and breadth.

Ans: Let the length of the park be $\text{x m}$ and breadth of the park be $\text{x m}$.

Thus, $\text{Perimeter=2}\left( \text{x+y} \right)$.

$\text{2}\left( \text{x+y} \right)\text{=80}$

$\Rightarrow \text{x+y=40}$

$\Rightarrow \text{y=40-x}$.

Now, $\text{Area=x }\!\!\times\!\!\text{ y}$.

Substituting value of y.

$\text{Area=x}\left( \text{40-x} \right)$

$\text{x}\left( \text{40-x} \right)\text{=400}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-40x+400=0}$

So, $\text{a=1}$, $\text{b=-40}$, $\text{c=400}$.

Discriminant $\text{=}{{\left( \text{-40} \right)}^{\text{2}}}\text{-4}\left( \text{1} \right)\left( 400 \right)$

$\text{=1600-1600}$

Therefore, there is equal real roots for the given equation and hence, this situation is possible.

$\dfrac{\text{-b}}{\text{2a}}\text{=}\dfrac{\text{-}\left( -40 \right)}{2}$

$\text{=}\dfrac{\text{40}}{2}$

\[\text{=20}\]

Therefore, length of park is $\text{x=20m}$ .

And breadth of park be $\text{y=}\left( \text{40-20} \right)\text{m}$ i.e., $\text{y=20m}$.

Important Points from NCERT Class 10 Quadratic Equations

A quadratic equation can be represented as:

ax 2 + bx + c = 0

Where x is the variable of the equation and a, b and c are the real numbers. Also, a≠0.

The nature of roots of a quadratic equation ax 2 + bx + c = 0 can be find as:

A real number α be root of quadratic equations ax 2 + bx + c = 0 if and only if 

aα 2 + bα + c = 0.

Quadratic equations are very important in real-life situations. Learn all the concepts deeply and understand each topic conceptually. And, now let us solve questions related to quadratic equations.

Maths Class 10 Quadratic Equations Mind Map

Relation between the zeroes of a quadratic equation and the coefficient of a quadratic equation.

If α and β are zeroes of the quadratic equation $ax^2 + bx + c = 0$, where a, b, and c are real numbers and a ≠ 0, then

$\alpha + \beta = -\dfrac{b}{a}$

$\text{sum of zeros} = -\dfrac{\text{coefficient of x}}{\text{coefficient of }x^2}$

$\alpha \beta = \dfrac{c}{a}$

$\text{product of zeros} = -\dfrac{\text{constant term}}{\text{coefficient of }x^2}$

Methods of Solving a Quadratic Equation

The following are the methods that are used to solve quadratic equations:

(i) Factorization; (ii) Completing the Square; (iii) Quadratic Formula

Methods of Factorization

In this method, we find the roots of a quadratic equation $(ax^2 + bx + c = 0)$ by factorizing LHS into two linear factors and equating each factor to zero, e.g., $6x^2 - x - 2 = 0$ $\Rightarrow 6x^2 + 3x - 4x - 2 = 0$ …(i) $\Rightarrow 3x (2x + 1) - 2(2x + 1) = 0$ $\Rightarrow (3x - 2) (2x + 1) = 0$ $\Rightarrow 3x - 2 = 0$ or $2x + 1 = 0$

Therefore $x = \dfrac{2}{3}$ or $x = -\dfrac{-1}{2}$

Method of Completing the Square

This is the method of converting the LHS of a quadratic equation that is not a perfect square into the sum or difference of a perfect square and a constant by adding and subtracting the terms.

Quadratic Formula

Consider a quadratic equation: ax2 + bx + c = 0. If b2 – 4ac ≥ 0, then the roots of the above equation are given by:

$x = -\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Overview of Deleted Syllabus for CBSE Class 10 Maths Chapter 4 Quadratic Equation

Class 10 Maths Chapter 4: Exercise Breakdown

NCERT Solutions for Class 10 Maths Chapter 4 - provides a comprehensive guide to understanding quadratic equations. This chapter is important for students because it teaches topics that are fundamental to advanced mathematics. The solutions describe how to solve quadratic equations, apply the quadratic formula, and investigate the nature of the roots using the discriminant. For effective exam preparation, focus on understanding formula derivation and applying the many types of problem-solving approaches described in this chapter. Last year, four to six questions from this area featured in the board exams, demonstrating its importance. These answers are precisely crafted to help students succeed by improving their problem-solving skills and conceptual understanding.

Other Study Materials of CBSE Class 10 Maths Quadratic Equation

Chapter-Specific NCERT Solutions for Class 10 Maths

Given below are the chapter-wise NCERT Solutions for Class 10 Maths . Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

FAQs on NCERT Solutions for Class 10 Maths Chapter 4 - Quadratic Equations

The subject’s experts take online classes available at Vedantu. They have lots of experience in their respective subjects. Along with this, they are working in their field with consistency. So, they have been faced with all kinds of problems and learned the tricks on how to come out from it.

In online classes, teachers share all their experiences with the students and teach some essential tricks. These tricks will be useful at the time of solving questions in the examination hall. Apart from the teaching concepts and formulas, they also motivate students and remove the fear of examinations from the student’s mind that is built up somewhere due to some kind of society’s pressure and other things.

2. Instead of Class 10th Maths Chapter 4, What is the Maths Syllabus for Class 10 for 202 4-25 ?

Here is the complete syllabus for Class 10 Mathematics Revised Syllabus 2024-25:-

Unit- I: Number Systems

Real Numbers

Unit II: Algebra

Polynomials

Pair of Linear Equations in Two Variables

Quadratic Equations

 Arithmetic Progressions

Unit III: Coordinate Geometry

Lines (In two-dimensions)

 Unit IV: Geomtry

Constructions

Unit V: Trigonometry

Introduction to Trigonometry

Trigonometric Identities

 Heights and Distances: Angle of elevation, Angle of Depression

Unit VI: Mensuration

Areas Related to Circles

Surface Areas and Volumes

Unit VII: Statistics and Probability

 Statistics

Probability

3. What is the Weightage for Class 10 Mathematics Unit-Wise?

Students who don’t know the weightage then they should go through the table given below. Here, we have mentioned the entire Class 10 Mathematics Unit-Wise Weightage for the knowledge of the students.

4. Mention the important concepts that you learn in NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations.

If you want to score 100 per cent marks in Class 10 Maths, you have to practice daily. Chapter 4 Quadratic Equations is an important chapter. Students can find NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations on Vedantu. There are five exercises in Chapter 4. Students can download the NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations to learn the important concepts that will help them understand the topic.

5. How to download Class 10 Maths Quadratic Equations NCERT Textbooks PDF?

Students can easily download Class 10 Maths Quadratic Equations NCERT textbooks PDF online. NCERT Solutions for Class 10 Maths Quadratic Equations are explained in an easy and simple language. Follow the given steps:

Click NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations.

Click on “Download PDF”.

Download and save it.

Students can use the PDF document without having an internet connection and can study Maths Quadratic Equations anytime. The NCERT Solutions give a clear understanding to them. 

6. What is the Quadratic formula Class 10th?

When a quadratic polynomial is equated to a constant, it forms a quadratic equation. An equation such as Ax = D, where Ax is a polynomial of degree two and D is a constant, forms a quadratic equation. The standard quadratic equation is a x 2 +bx+c=0 where a, b, and c are not equal to zero. You can refer to NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations to understand more about the topic. You can also download Vedantu’s app. All the resources are available free of cost. 

7. What are the important topics covered in NCERT Solutions Class 10 Maths Chapter 4?

Chapter 4 Class 10 Maths is based on Quadratic Equations. The chapter includes important concepts about quadratic equations. This chapter includes five exercises that explain the different concepts of quadratic equations. The following topics are covered:

Exercise 4.1- Introduction

Exercise 4.2- Quadratic Equations

Exercise 4.3- Solution of a Quadratic Equation by Factorisation

Exercise 4.4- Solution of a Quadratic Equation by Completing the Square

Exercise 4.5- Nature of Roots

8. How do you solve Quadratic Equations in Class 10?

If you want to learn how to solve quadratic equations in Class 10, you can refer to NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations. All the solutions are prepared by experts in an easy language. Students can understand the equations clearly. Students have to find the roots by using the quadratic formula. They can find the sum and product of both the roots. The method is simple and explained properly for easier understanding.

NCERT Solutions for Class 10 Maths

Ncert solutions for class 10.

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Case Study Class 10 Maths Questions

Table of Contents

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Download the app to get CBSE Sample Papers 2023-24, NCERT Solutions (Revised), Most Important Questions, Previous Year Question Bank, Mock Tests, and Detailed Notes.

Now, CBSE will ask only subjective questions in class 10 Maths case studies. But if you search over the internet or even check many books, you will get only MCQs in the class 10 Maths case study in the session 2022-23. It is not the correct pattern. Just beware of such misleading websites and books.

We advise you to visit CBSE official website ( cbseacademic.nic.in ) and go through class 10 model question papers . You will find that CBSE is asking only subjective questions under case study in class 10 Maths. We at myCBSEguide helping CBSE students for the past 15 years and are committed to providing the most authentic study material to our students.

Here, myCBSEguide is the only application that has the most relevant and updated study material for CBSE students as per the official curriculum document 2022 – 2023. You can download updated sample papers for class 10 maths .

First of all, we would like to clarify that class 10 maths case study questions are subjective and CBSE will not ask multiple-choice questions in case studies. So, you must download the myCBSEguide app to get updated model question papers having new pattern subjective case study questions for class 10 the mathematics year 2022-23.

Class 10 Maths has the following chapters.

• Real Numbers Case Study Question
• Polynomials Case Study Question
• Pair of Linear Equations in Two Variables Case Study Question
• Quadratic Equations Case Study Question
• Arithmetic Progressions Case Study Question
• Triangles Case Study Question
• Coordinate Geometry Case Study Question
• Introduction to Trigonometry Case Study Question
• Some Applications of Trigonometry Case Study Question
• Circles Case Study Question
• Area Related to Circles Case Study Question
• Surface Areas and Volumes Case Study Question
• Statistics Case Study Question
• Probability Case Study Question

Format of Maths Case-Based Questions

CBSE Class 10 Maths Case Study Questions will have one passage and four questions. As you know, CBSE has introduced Case Study Questions in class 10 and class 12 this year, the annual examination will have case-based questions in almost all major subjects. This article will help you to find sample questions based on case studies and model question papers for CBSE class 10 Board Exams.

Maths Case Study Question Paper 2023

Here is the marks distribution of the CBSE class 10 maths board exam question paper. CBSE may ask case study questions from any of the following chapters. However, Mensuration, statistics, probability and Algebra are some important chapters in this regard.

Case Study Question in Mathematics

Here are some examples of case study-based questions for class 10 Mathematics. To get more questions and model question papers for the 2021 examination, download myCBSEguide Mobile App .

Case Study Question – 1

In the month of April to June 2022, the exports of passenger cars from India increased by 26% in the corresponding quarter of 2021–22, as per a report. A car manufacturing company planned to produce 1800 cars in 4th year and 2600 cars in 8th year. Assuming that the production increases uniformly by a fixed number every year.

• Find the production in the 1 st year.
• Find the production in the 12 th year.
• Find the total production in first 10 years. OR In which year the total production will reach to 15000 cars?

Case Study Question – 2

In a GPS, The lines that run east-west are known as lines of latitude, and the lines running north-south are known as lines of longitude. The latitude and the longitude of a place are its coordinates and the distance formula is used to find the distance between two places. The distance between two parallel lines is approximately 150 km. A family from Uttar Pradesh planned a round trip from Lucknow (L) to Puri (P) via Bhuj (B) and Nashik (N) as shown in the given figure below.

• Find the distance between Lucknow (L) to Bhuj(B).
• If Kota (K), internally divide the line segment joining Lucknow (L) to Bhuj (B) into 3 : 2 then find the coordinate of Kota (K).
• Name the type of triangle formed by the places Lucknow (L), Nashik (N) and Puri (P) OR Find a place (point) on the longitude (y-axis) which is equidistant from the points Lucknow (L) and Puri (P).

Case Study Question – 3

• Find the distance PA.
• Find the distance PB
• Find the width AB of the river. OR Find the height BQ if the angle of the elevation from P to Q be 30 o .

Case Study Question – 4

• What is the length of the line segment joining points B and F?
• The centre ‘Z’ of the figure will be the point of intersection of the diagonals of quadrilateral WXOP. Then what are the coordinates of Z?
• What are the coordinates of the point on y axis equidistant from A and G? OR What is the area of area of Trapezium AFGH?

Case Study Question – 5

The school auditorium was to be constructed to accommodate at least 1500 people. The chairs are to be placed in concentric circular arrangement in such a way that each succeeding circular row has 10 seats more than the previous one.

• If the first circular row has 30 seats, how many seats will be there in the 10th row?
• For 1500 seats in the auditorium, how many rows need to be there? OR If 1500 seats are to be arranged in the auditorium, how many seats are still left to be put after 10 th row?
• If there were 17 rows in the auditorium, how many seats will be there in the middle row?

Case Study Question – 6

• Draw a neat labelled figure to show the above situation diagrammatically.

• What is the speed of the plane in km/hr.

More Case Study Questions

We have class 10 maths case study questions in every chapter. You can download them as PDFs from the myCBSEguide App or from our free student dashboard .

As you know CBSE has reduced the syllabus this year, you should be careful while downloading these case study questions from the internet. You may get outdated or irrelevant questions there. It will not only be a waste of time but also lead to confusion.

Here, myCBSEguide is the most authentic learning app for CBSE students that is providing you up to date study material. You can download the myCBSEguide app and get access to 100+ case study questions for class 10 Maths.

How to Solve Case-Based Questions?

Questions based on a given case study are normally taken from real-life situations. These are certainly related to the concepts provided in the textbook but the plot of the question is always based on a day-to-day life problem. There will be all subjective-type questions in the case study. You should answer the case-based questions to the point.

What are Class 10 competency-based questions?

Competency-based questions are questions that are based on real-life situations. Case study questions are a type of competency-based questions. There may be multiple ways to assess the competencies. The case study is assumed to be one of the best methods to evaluate competencies. In class 10 maths, you will find 1-2 case study questions. We advise you to read the passage carefully before answering the questions.

Case Study Questions in Maths Question Paper

CBSE has released new model question papers for annual examinations. myCBSEguide App has also created many model papers based on the new format (reduced syllabus) for the current session and uploaded them to myCBSEguide App. We advise all the students to download the myCBSEguide app and practice case study questions for class 10 maths as much as possible.

Case Studies on CBSE’s Official Website

CBSE has uploaded many case study questions on class 10 maths. You can download them from CBSE Official Website for free. Here you will find around 40-50 case study questions in PDF format for CBSE 10th class.

10 Maths Case Studies in myCBSEguide App

You can also download chapter-wise case study questions for class 10 maths from the myCBSEguide app. These class 10 case-based questions are prepared by our team of expert teachers. We have kept the new reduced syllabus in mind while creating these case-based questions. So, you will get the updated questions only.

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• CBSE Reduced Syllabus Class 10 (2020-21)
• Class 10 Maths Basic Sample Paper 2024
• How to Revise CBSE Class 10 Maths in 3 Days
• CBSE Practice Papers 2023
• Class 10 Maths Sample Papers 2024
• Competency Based Learning in CBSE Schools

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NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

NCERT Solutions Class 10 Maths Chapter 4 Quadratic Equations are provided here to help the students of CBSE class 10. Our expert teachers prepared all these solutions as per the latest CBSE syllabus and guidelines. In this chapter, we have discussed how to find the solution of a quadratic equation by – factorisation, completing the square method in details. CBSE Class 10 Maths solutions provide a detailed and step-wise explanation of each answer to the questions given in the exercises of NCERT books.

CBSE Class 10 Maths Chapter 4 Quadratic Equations Solutions

Below we have given the answers to all the questions present in Quadratic Equations in our NCERT Solutions for Class 10 Maths chapter 4. In this lesson, students are introduced to a lot of important concepts that will be useful for those who wish to pursue mathematics as a subject in their future classes. Based on these solutions, students can prepare for their upcoming Board Exams. These solutions are helpful as the syllabus covered here follows NCERT guidelines.

NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.1

NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.2

NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.3

NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.4

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Case Study Class 10 Maths Questions and Answers (Download PDF)

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Case Study Class 10 Maths

If you are looking for the CBSE Case Study class 10 Maths in PDF, then you are in the right place. CBSE 10th Class Case Study for the Maths Subject is available here on this website. These Case studies can help the students to solve the different types of questions that are based on the case study or passage.

CBSE Board will be asking case study questions based on Maths subjects in the upcoming board exams. Thus, it becomes an essential resource to study. 

The Case Study Class 10 Maths Questions cover a wide range of chapters from the subject. Students willing to score good marks in their board exams can use it to practice questions during the exam preparation. The questions are highly interactive and it allows students to use their thoughts and skills to solve the given Case study questions.

Download Class 10 Maths Case Study Questions and Answers PDF (Passage Based)

Download links of class 10 Maths Case Study questions and answers pdf is given on this website. Students can download them for free of cost because it is going to help them to practice a variety of questions from the exam perspective.

Case Study questions class 10 Maths include all chapters wise questions. A few passages are given in the case study PDF of Maths. Students can download them to read and solve the relevant questions that are given in the passage.

Students are advised to access Case Study questions class 10 Maths CBSE chapter wise PDF and learn how to easily solve questions. For gaining the basic knowledge students can refer to the NCERT Class 10th Textbooks. After gaining the basic information students can easily solve the Case Study class 10 Maths questions.

Case Study Questions Class 10 Maths Chapter 1 Real Numbers

Case Study Questions Class 10 Maths Chapter 2 Polynomials

Case Study Questions Class 10 Maths Chapter 3 Pair of Equations in Two Variables

Case Study Questions Class 10 Maths Chapter 4 Quadratic Equations

Case Study Questions Class 10 Maths Chapter 5 Arithmetic Progressions

Case Study Questions Class 10 Maths Chapter 6 Triangles

Case Study Questions Class 10 Maths Chapter 7 Coordinate Geometry

Case Study Questions Class 10 Maths Chapter 8. Introduction to Trigonometry

Case Study Questions Class 10 Maths Chapter 9 Some Applications of Trigonometry

Case Study Questions Class 10 Maths Chapter 10 Circles

Case Study Questions Class 10 Maths Chapter 12 Areas Related to Circles

Case Study Questions Class 10 Maths Chapter 13 Surface Areas & Volumes

Case Study Questions Class 10 Maths Chapter 14 Statistics

Case Study Questions Class 10 Maths Chapter 15 Probability

How to Solve Case Study Based Questions Class 10 Maths?

In order to solve the Case Study Based Questions Class 10 Maths students are needed to observe or analyse the given information or data. Students willing to solve Case Study Based Questions are required to read the passage carefully and then solve them. 

While solving the class 10 Maths Case Study questions, the ideal way is to highlight the key information or given data. Because, later it will ease them to write the final answers. 

Case Study class 10 Maths consists of 4 to 5 questions that should be answered in MCQ manner. While answering the MCQs of Case Study, students are required to read the paragraph as they can get some clue in between related to the topics discussed.

Also, before solving the Case study type questions it is ideal to use the CBSE Syllabus to brush up the previous learnings.

Features Of Class 10 Maths Case Study Questions And Answers Pdf

Students referring to the Class 10 Maths Case Study Questions And Answers Pdf from Selfstudys will find these features:-

• Accurate answers of all the Case-based questions given in the PDF.
• Case Study class 10 Maths solutions are prepared by subject experts referring to the CBSE Syllabus of class 10.
• Free to download in Portable Document Format (PDF) so that students can study without having access to the internet.

Benefits of Using CBSE Class 10 Maths Case Study Questions and Answers

Since, CBSE Class 10 Maths Case Study Questions and Answers are prepared by our maths experts referring to the CBSE Class 10 Syllabus, it provided benefits in various way:-

• Case study class 10 maths helps in exam preparation since, CBSE Class 10 Question Papers contain case-based questions.
• It allows students to utilise their learning to solve real life problems.
• Solving case study questions class 10 maths helps students in developing their observation skills.
• Those students who solve Case Study Class 10 Maths on a regular basis become extremely good at answering normal formula based maths questions.
• By using class 10 Maths Case Study questions and answers pdf, students focus more on Selfstudys instead of wasting their valuable time.
• With the help of given solutions students learn to solve all Case Study questions class 10 Maths CBSE chapter wise pdf regardless of its difficulty level.

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• NCERT Solutions for Class 10 Maths
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Class 10 Maths Extra Questions Chapter 4 Quadratic Equations

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CBSE Class 10 Maths Competency-Based Questions With Answer Key 2024-25: Chapter 6 Triangles Free PDF Download

Cbse class 10 maths chapter 6 practice questions 2025: students check and download the cbse class 10 chapter 6 triangles competency-focused practice questions along with answer key for 2024-25. .

CBSE 2024-25 Competency Based Questions With Answers: The Central Board of Secondary Education (CBSE) has designed and released the competency‐focused practice questions for students of classes 10 and 12. Students can practice questions ranging from multiple choice questions to case studies. CBSE has made available the practice questions on its website, cbse.academic.nic.in. This article covers the CBSE Class 10 Maths Chapter 6 Triangles competency-focused questions. 

These assessments contain high quality questions and concepts that are important for learning. When students will solve the competency-based questions, they will engage in learning by doing. The questions are designed to test the understanding of students and remove any gaps that may exist in the learning process. Competency-based practice questions follow the principle of learning with understanding. 

CBSE Class 10 Maths Chapter 6 Competency Based-Questions 

Tips to Prepare for CBSE Class 10 Mathematics Examination

1. Students must first familiarize themselves with the syllabus so as to understand which topics carry more weightage. Ensure that you prepare according to the latest syllabus only. 

2. Mathematics is a subject that demands regular practice. It is important that students practice as much as they can. Practice the questions given at the end of each chapter. Note the points where you get stuck during solving questions. 

3. Students should prepare an effective study plan and give time to all topics on the basis of their strengths and weaknesses. Don’t let your bias towards your favourite chapter affect the time-table. The schedule should also have time for revision. 

4. Students should solve the problems from CBSE Class 10 subject textbook to clear the concepts and strengthen understanding. While solving problems, students should make it a habit to show rough work clearly which will help them in the examination as well. 

CBSE Video Courses for Class 10 Students 

Class 10 students can study effectively for the exams with the help of video courses prepared by the subject matter experts. These video courses will explain the concepts in a simple and interactive manner which will help learners to understand clearly. 

CBSE Class 10 Video Courses 

Also, check

CBSE Class 10 Syllabus 2024-25: Download Latest and Revised FREE PDFs

NCERT Books for Class 10 All Subjects PDF (2024-25)

NCERT Solutions for Class 10 (2024-2025) All Subjects & Chapters: Download in PDF

CBSE Class 10 Maths Mind Maps for Quick Revision

CBSE Class 10 Maths MCQs: Important for 2024-2025 Exams, Download Chapter-Wise PDF

Pythagoras Theorem: Definition, Formula, Proof, Examples and Applications

Polynomial Equations: Concepts, Definition, Formula and Types

Relations and Functions

Probability Notes: Definition, Types, Formulas, Applications and More

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