NCERT Solutions For Class 10 Maths

Chapter 1 - Real Numbers

In Class 10, Chapter 1 introduces real and irrational numbers. It begins with Euclid’s Division Lemma, stating that if you have two positive numbers, a and b, you can divide a by b and get a unique quotient q and remainder r. Euclid’s Division algorithm uses this idea to find the Highest Common Factor (HCF) of two positive numbers. Then, the chapter talks about the Fundamental Theorem of Arithmetic, which helps find both the Lowest Common Multiple (LCM) and HCF of two positive numbers. Lastly, it covers irrational and rational numbers, along with their decimal forms, using this theorem.

Topics Covered in Class 10 Maths Chapter 1 Real Numbers :

• Real Numbers

• Euclid’s Division Lemma

• Fundamental Theorem of Arithmetic

• Irrational Numbers

• Rational Numbers

Important Steps:

The Fundamental Theorem of Arithmetic talks about how every number can be expressed as a unique product of prime numbers. To understand this, we look at examples and reasons. We also prove that $sqrt{2}, \sqrt{3},$ and $\sqrt{5}$ are irrational.

To find the highest common factor (HCF) of two positive numbers, follow these steps:

1. Use Euclid’s division method to divide the larger number by the smaller one. This gives a quotient and a remainder.

2. If the remainder is zero, the smaller number is the HCF. If not, repeat the division with the smaller number and the remainder.

3. Keep repeating this process until the remainder becomes zero. The divisor at this stage will be the HCF. This works because the HCF of two numbers is the same as the HCF of the smaller number and the remainder.

Students can access extra study materials for Real Numbers on Vedantu. These resources are available for download, offering additional support for your studies.

• Real Numbers Important Questions

• Real Numbers Important Formulas

• Real Numbers Revision Notes

• Real Numbers NCERT Exemplar Solutions

• Real Numbers RD Sharma Solutions

Chapter 2 - Polynomials

The Polynomials chapter covers different types of polynomials like linear, quadratic, and cubic. It starts by explaining what the degree of a polynomial means. There are four exercises in total, one of which is optional.

• Exercise 2.1 focuses on figuring out the number of zeroes using graphs. You'll need to understand what zeroes of a polynomial mean in terms of geometry.

• Exercise 2.2 is about the connection between zeroes and coefficients of a polynomial. You'll be finding zeroes of quadratic polynomials and sometimes even figuring out the polynomial itself.

• Exercise 2.3 introduces the division algorithm concept, with related questions.

• The optional exercise, 2.4, combines questions from all the concepts in Chapter 2.

Topics Covered in Class 10 Polynomials

• Zeros of a polynomial. 

• Relationship between zeroes 

• Coefficients of quadratic polynomials only.

Important Steps:

Arrange in Standard Form: First, we line up the terms of the numbers we're dividing so that the highest powers come first. This is called the standard form.

• Step 1 - To get the first term of the quotient, divide the dividend's highest degree term by the divisor's highest degree term. Then, complete the dividing procedure.

• Step 2 - To calculate the second term of the quotient, divide the highest degree term of the new dividend by the divisor's highest degree term. Continue the dividing process.

• Step 3 - Now, the remainder's degree is smaller than the divisor's. So we can't prolong the separation any further.

Dividend = Divisor × Quotient + Remainder. What we are doing here is using an approach similar to Euclid's division algorithm, which you discussed in Chapter 1.

This states that.

If p(x) and g(x) are two polynomials with g(x) ≠ 0, we may find polynomials q(x) and r(x) such that

P(x) = g(x) × q(x) + r(x).

Where r(x) = 0 or  degree of r(x) <  degree of g(x).

This conclusion is referred to as the Division Algorithm for Polynomials.

Students can access extra study materials for Polynomials on Vedantu. These resources are available for download, offering additional support for your studies.

• Polynomials Important Questions

• Polynomials Important Formulas

• Polynomials Revision Notes

• Polynomials NCERT Exemplar Solutions

• Polynomials RD Sharma Solutions

Chapter 3 - Pair of Linear Equations in Two Variables

Chapter 3 is about the linear equations in two variables. There are 7 exercises in this section. It explains what two-variable linear equations are and how to solve them.

• Exercise 3.1, representing a problem or situation algebraically and graphically is taught. 

• Exercise 3.2, depicts the methods of solving linear equations by graphical method.

• Exercises 3.3, 3.4, 3.5 and 3.6 explain different methods for solving pairs of linear equations: algebraic, cross-multiplication, elimination, and substitution. 

• Exercise 3.7  is an optional exercise as it has questions from all the previous exercises.

Topics Covered in Class 10 Pair of Linear Equations in Two Variables

• Pair of linear equations in two variables and graphical method of their solution, consistency/inconsistency. 

• Algebraic conditions for number of solutions. 

• Solution of a pair of linear equations in two variables algebraically – by substitution and by elimination. 

• Simple situational problems. 

• Simple problems on equations reducible to linear equations.

Important Steps:

A pair of two-variable linear equations and a graphical technique for solving them, as well as consistency and inconsistency. Number of solutions is subject to algebraic conditions. Solve a pair of linear equations in two variables algebraically - using substitution and elimination. Simple situational problems.

The equations typically look like this:

The general form for a pair of linear equations in two variables, x and y, is

$a_1 x + b_1 y + c_1 = 0$

and $a_2 x + b_2 y + c_2 = 0$

where $a_1, b_1, c_1, a_2, b_2, c_2$ are real numbers

To solve them algebraically, you can use substitution or elimination methods.

Here are some key formulas to remember:

$a_1^2 + b_1^2 \neq 0$

$a_2^2 + b_2^2 \neq 0$

Students can access extra study materials for Pair of Linear Equations in Two Variables on Vedantu. These resources are available for download, offering additional support for your studies.

• Pair of Linear Equations in Two Variables Important Questions

• Pair of Linear Equations in Two Variables Important Formulas

• Pair of Linear Equations in Two Variables Revision Notes

• Pair of Linear Equations in Two Variables NCERT Exemplar Solutions

• Pair of Linear Equations in Two Variables RD Sharma Solutions

Chapter 4 - Quadratic Equations

Chapter 4 focuses on quadratic equations. This chapter defines a quadratic equation and discusses several strategies for solving them. It also goes over the factorization approach for solving quadratic equations and the square method. The chapter describes the links between discriminants and the nature of roots. This chapter also provides instances of real-world challenges that have been solved. The chapter concludes with a discussion of determining the roots of a quadratic equation $ax^2 + bx + c = 0$.

• If $b^2 - 4ac > 0$, the equation has two unique real roots.

• Find two equal roots of $b^2 - 4ac = 0$.

• No genuine roots if $b^2 - 4ac < 0$.

Topics Covered in Class 10 Quadratic Equations

• Standard form of a quadratic equation ax2 + bx + c = 0, (a ≠ 0). 

• Solutions of quadratic equations (only real roots) by factorization, and by using the quadratic formulas. 

• Relationship between discriminant and nature of roots. 

• Situational problems based on quadratic equations.

Important Steps:

The conventional form of a quadratic equation is $ax^2 + bx + c = 0$ where a ≠ 0. Quadratic equations (only real roots) can be solved by factoring and applying the quadratic formula. Relationship between the discriminant and the nature of roots.

Situational questions based on quadratic equations relating to daily activities have to be included.

If $b^2 - 4ac > 0$, we obtain two unique real roots.   

$-\dfrac{b}{2a}+\dfrac{\sqrt{b^2-4ac}}{2a}$ and $-\dfrac{b}{2a}-\dfrac{\sqrt{b^2-4ac}}{2a}$

If $b^2 - 4ac = 0$, then 

$x=-\dfrac{b}{2a}\pm 0$

$x=-\dfrac{b}{2a}\pm 0$

So, the roots of the equation $ax^2 + bx + c = 0$ are both $-\dfrac{b}{2a}$.

As a result, we may conclude that the quadratic equation $ax^2 + bx + c = 0$ has two equal real roots in this situation.

If $b^2 - 4ac < 0$, there is no real number whose square equals $b^2 - 4ac$. As a result, the above quadratic equation does not have any real roots.

The discriminant of the quadratic equation $ax^2 + bx + c = 0$ is defined as $b^2 - 4ac$.

So, a quadratic equation $ax^2 + bx + c = 0$ has

(i) two distinct real roots, if $b^2 - 4ac > 0$,

(ii) two equal real roots, if $b^2 - 4ac = 0$,

(iii) no real roots, if $b^2 - 4ac < 0$.

Students can access extra study materials for Quadratic Equations on Vedantu. These resources are available for download, offering additional support for your studies.

• Quadratic Equations Important Questions

• Quadratic Equations Important Formulas

• Quadratic Equations Revision Notes

• Quadratic Equations NCERT Exemplar Solutions

• Quadratic Equations RD Sharma Solutions

Chapter 5 - Arithmetic Progressions

This chapter introduces a new topic to the students - Arithmetic Progression, commonly called AP. The chapter constitutes a total of 4 exercises. Learn what AP is, the derivation of the nth term, finding the sum of first n terms of the AP and solving the real-life problems using AP in this chapter.

• Exercise 5.1 of the chapter teaches how to represent a problem or situation in the form of AP, how to find the first term and difference of the AP and to find whether the given series is AP or not. 

• Exercise 5.2 explains how to find the nth term of the AP using the formula. 

$a_n = a + (n-1) d$

• Exercise 5.3 describes how to find the sum of the first n terms of AP and contains questions related to the topic. 

• Exercises 5.4 have questions related to the topics taught in the chapter.

Topics Covered in Class 10 Arithmetic Progressions

• Motivation for studying Arithmetic Progression Derivation of the nth term and sum of the first n terms of A.P. and their application in solving daily life problems.

Important Formulas

The sequence $a_1, a_2, a_3, a_4, a_5, a_6,…$ of AP terms is written as a, a+d, a+2d, a+3d, a+4d, a+5d, …., nth term, where $a_n$ is the initial term, d is the common difference between each term.

The nth term for arithmetic progression is given as,

$a + (n-1) d$

Sum of the first n terms in Arithmetic Progression.

$S_n = \dfrac{n}{2}[2a + (n-1) d]$.

Students can access extra study materials for Arithmetic Progressions on Vedantu. These resources are available for download, offering additional support for your studies.

• Arithmetic Progressions Important Questions

• Arithmetic Progressions Revision Notes

• Arithmetic Progressions NCERT Exemplar Solutions

• Arithmetic Progressions RD Sharma Solutions

Chapter 6 - Triangles

Chapter 6 of class 10 gives details about the triangles. The chapter gives details about the figures with the same shape but different sizes. It explains the similarity of the triangles, theorems associated with the similarity of triangles and the concept of congruent triangles. Further, theorems related to areas of triangles, the Pythagoras theorem and the converse of Pythagoras theorem is explained.

Topics Covered in Class 10 Triangles

• Concept of a similar and congruent figure

• Theorems related to the similarity of triangles

• Areas of similar triangles

• Pythagoras Theorem and converse of Pythagoras Theorem

Definitions, Examples, and Counterexamples for Similar Triangles.

1. If a line is drawn parallel to one side of a triangle and intersects the other two sides at separate spots, the remaining two sides are split in the same ratio - Prove

2. A line that splits two sides of a triangle in the same ratio is parallel to the third side - Motivate

3. If the angles of two triangles are equal, the sides are proportionate, and the triangles are identical - Motivate

4. If the sides of two triangles are proportionate, their angles are equal, and the triangles are identical - Motivate

5. If one angle of a triangle equals one angle of another triangle, and the sides that include these angles are proportionate, the two triangles are comparable - Motivate

Important Theorems:

• Theorem 6.1: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

• Theorem 6.2: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

• Theorem 6.3: If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion), and hence the two triangles are similar.

• Theorem 6.4: If in two triangles, the sides of one triangle are proportional to (i.e., in the same ratio of ) the sides of the other triangle, then their corresponding angles are equal, and hence the two triangles are similar.

• Theorem 6.5: If one angle of a triangle is equal to one angle of the other triangle and the sides, including these angles, are proportional, then the two triangles are similar.

• Theorem 6.6: The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

• Theorem 6.7: If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

• Theorem 6.8: In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

• Theorem 6.9: In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

Students can access extra study materials for Triangles on Vedantu. These resources are available for download, offering additional support for your studies.

• Triangles Important Questions

• Triangles Revision Notes

• Triangles NCERT Exemplar Solutions

• Triangles RD Sharma Solutions

Chapter 7 - Coordinate Geometry

Chapter 7 introduces coordinate geometry, which is the mathematical process of identifying a given location using an ordered pair of integers. Coordinate or cartesian geometry is used to calculate the distance between two places given their coordinates. The notion of calculating the area of a triangle formed by three specified points. In addition, the chapter explains how to get the coordinates of the point that splits a line segment connecting two specified points in a certain ratio. This chapter of NCERT Solutions teaches students about the distance formula, the section formula, and the area of triangles.

Topics Covered in Class 10 Coordinate Geometry

• Concepts of coordinate geometry

• Graphs of linear equations

• Distance formula 

• Section formula (Internal division)

Important Formulas:

Distance Formula: $PQ = \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$

Section Formula: $m_1: m_2=\left(\frac{m_1 x_2+m_2 x_1}{m_1+m_2}, \frac{m_1 y_2+m_2 y_1}{m_1+m_2}\right)$

Area of Triangle: = $\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]$

Students can access extra study materials for Coordinate Geometry on Vedantu. These resources are available for download, offering additional support for your studies.

• Coordinate Geometry Important Questions

• Coordinate Geometry Revision Notes

• Coordinate Geometry NCERT Exemplar Solutions

• Coordinate Geometry RD Sharma Solutions

Chapter 8 - Introduction to Trigonometry

Chapter 8 of NCERT Solutions introduces trigonometry to the students. Trigonometry is the study of the ratio of right triangles with respect to the acute angles, which are known as trigonometric ratios of the angles. Know the trigonometric ratios for angles $0^{\circ}$ and $90^{\circ}$ and the values of $30^{\circ}, 45^{\circ}$ and $60^{\circ}$. Students will also learn to calculate trigonometric ratios for the given angles and also a few trigonometric identities. The introductory formulas of trigonometry help to have a strong grip over the basics. In addition to the formulas, the relationship between the ratios and angles is also explained in the chapter.

Topics Covered in Class 10 Introduction to Trigonometry

• Trigonometric ratios of an acute angle of a right-angled triangle. 

• Proof of their existence. 

• Values of the trigonometric ratios. 

• Relationships between the ratios.

Important Formulas:

Trigonometry Maths formulae for Class 10 include the three major functions Sine, Cosine, and Tangent for a right-angle triangle. Let ABC be a right-angled triangle with ∠θ at point B.

$\sin \theta = \dfrac{\text{Side opposite to angle }\theta}{\text{Hypotenuse}} = \frac{\text { Perpendicular }}{\text { Hypotenuse }}$

$\cos \theta = \dfrac{\text{Adjacent side to angle }\theta}{\text{Hypotenuse}} = \dfrac{\text{Base}}{\text { Hypotenuse }}$

$\tan \theta = \dfrac{\text { Side opposite to angle } \theta }{\text { Adjacent side to angle } \theta}$

$\sec \theta = \frac{1}{\cos \theta}$

$\cot \theta = \frac{1}{\tan \theta}$

$\text{cosec} \theta = \frac{1}{\sin \theta}$

$\tan \theta = \frac{\sin \theta}{\cos \theta}$

Trigonometry Table

Trigonometric Ratios of Complementary Angles

$\sin (90^\circ - A) = \cos A$

$\cos (90^\circ - A) = \sin A$

$\tan (90^\circ - A) = \cot A$

$\cot (90^\circ - A) = \tan A$

$\sec (90^\circ - A) = \text{cosec} A$

$\text{cosec} (90^\circ - A) = \sec A$

$\sin^2 A + \cos^2 A = 1$

$\sec^2 A - \tan^2 A = 1$ for $0^\circ \leq  A < 90^\circ$

$\text{cosec}^2 A = 1 + \cot^2 A$ for $0^\circ < A \leq  90^\circ$

Students can access extra study materials for Introduction to Trigonometry on Vedantu. These resources are available for download, offering additional support for your studies.

• Introduction to Trigonometry Important Questions

• Introduction to Trigonometry Revision Notes

• Introduction to Trigonometry & Its Equations NCERT Exemplar Solutions

Chapter 9 - Some Applications of Trigonometry

Chapter 9 is the continuation of Chapter 8 where the students will learn about the application of trigonometry. This chapter helps to understand how trigonometry is applied to our everyday life to find out the height and distance of various objects without measuring them, how it is used in navigation, map production, and calculating an island's location relative to longitudes and latitudes. They will learn about the terms line of sight, angle of elevation, and angle of depression.

Topics Covered in Class 10 Maths Chapter 9 - Some Applications of Trigonometry:

In Chapter 9, we delve into "Some Applications of Trigonometry" with a focus on "Heights and Distances." This involves exploring angles of elevation and depression, specifically at 30°, 45°, and 60°.

Let's understand a few key points:

• Line of Sight: This is the line drawn from your eye to the point you're looking at.

• Angle of Elevation: When the point you're viewing is above your eye level, like when you tilt your head up.

• Angle of Depression: When the point you're viewing is below your eye level, like when you lower your head.

To apply these concepts practically to solve problems related to heights and distances, consider three key factors:

1. The distance from the observer (DE) to the base of the object (minar).

2. The angle of elevation (∠BAC) from the observer to the top of the object.

3. The height of the observer (AE).

Assuming you know these three conditions, you can determine the height of the object (minar). In the figure, we observe that CD equals CB plus BD. Here, BD is equivalent to AE, which represents the height of the observer.

To find BC, we utilize trigonometric ratios involving ∠BAC or ∠A in triangle ABC. Since BC is the opposite side concerning the known ∠A, our options are narrowed down to using either tan A or cot A, as these ratios involve AB and BC.

Therefore, we can use tan A = BC/AB or cot A = AB/BC to find BC. Adding AE to BC yields the height of the minar.

Students can access extra study materials for Some Applications of Trigonometry on Vedantu. These resources are available for download, offering additional support for your studies.

• Some Applications of Trigonometry Important Questions

• Some Applications of Trigonometry Revision Notes

• Some Applications of Trigonometry RD Sharma Solutions

Chapter 10 - Circles

You have learned about circles and related terms like chords, segments, and arcs previously. Now, we are going to learn situations where a circle and a line are on the same plane. This chapter focuses on understanding Tangents to a Circle and the Number of Tangents from a Point on a Circle.

Prove the Tangent to a Circle at Point of Contact

1. The tangent at any point of a circle is perpendicular to the radius through the point of contact.

2. The lengths of tangents drawn from an external point to a circle are equal.

Theorems

Tangent to a circle at the point of contact

1. The tangent at any point of a circle is perpendicular to the radius through the point of contact.

2. The lengths of tangents drawn from an external point to a circle are equal.

3. Theorem 10.1: The tangent at any point of a circle is perpendicular to the radius through the point of contact.

4. Theorem 10.2: The lengths of tangents drawn from an external point to a circle are equal.

Case-based Theorem

Number of Tangents from a Point on a Circle

• Case 1: There is no tangent to a circle passing through a point lying inside the circle.

• Case 2: There is one and only one tangent to a circle passing through a point lying on the circle.

• Case 3: There are exactly two tangents to a circle through a point lying outside the circle.

Topics Covered in Class 10 Circles

• Various terms related to a circle such as a chord, segment.

• Important Theorems

• Tangent to a Circle.

• Number of Tangents from a Point on a Circle.

Students can access extra study materials for Circles on Vedantu. These resources are available for download, offering additional support for your studies.

• Circles Important Questions

• Circles Revision Notes

• Circles NCERT Exemplar Solutions

• Circles NCERT RD Sharma Solutions

Chapter 11 - Areas Related to Circles

As the name suggests, the chapter is about the perimeter and area of a circle. The concepts of finding the area of a sector and segment of a circle are clarified here. This chapter further explains how to find the area of the figures that contains a circle or part of a circle.

Topics Covered in Class 10 Areas Related to Circles

• Calculate the area of a circle

• Area of sectors and segments of a circle. 

• Problems based on areas and perimeter.

• Circumference of the above plane figures.

(When computing the area of a circle segment, only the central angles of 60°, 90°, and 120° should be considered.)

Important formulas from within the chapter for the students to practice–

• Circumference = $2 \pi r$

• Area of the circle = $\pi r^2$

• Area of the sector of angle $\theta = \dfrac{\pi}{360} \times \pi r^2$

• Length of an arc of a sector of angle $\theta = \dfrac{\pi}{360} \times 2 \pi r$ where r is the radius of the circle

Students can access extra study materials for Area Related to Circles on Vedantu. These resources are available for download, offering additional support for your studies.

• Areas Related to Circles Important Questions

• Areas Related to Circles Revision Notes

• Areas Related to Circles NCERT Exemplar Solutions

• Areas Related to Circles RD Sharma Solutions

Chapter 12 - Surface Areas and Volumes

Chapter 12 is about the surface areas and volumes of different objects and shapes. In the five Exercises of this chapter

• Exercise 12.1 consists of questions on calculating the surface area of an object made by combining any two of the fundamental solids, namely cuboid, cone, cylinder, sphere, and hemisphere. 

• Exercise 12.2, questions are focused on determining the volume of objects generated by merging any two cuboids, cones, cylinders, spheres, or hemispheres. 

• Exercise 12.3 addresses the questions in which a solid is transformed from one shape to another.

• Exercise 12.4 requires you to calculate the volume, curved surface area, and total surface area of a cone's frustum. The final exercise is optional and consists of high-level questions covering all of the chapter's subjects.

Topics Covered in Class 10 Surface Areas and Volumes

• Surface areas and volumes of combinations of any two figures like cubes, cuboids, spheres, hemispheres and right circular cylinders/cones.

• Problems that involve converting one type of metallic solid into another and other mixed problems.

Important Formulas Used for Calculating Surface Areas and Volumes of Different Objects

• TSA of new solid = CSA of one hemisphere + CSA of cylinder + CSA of other hemispheres

• Diameter of sphere = 2r

• The surface area of sphere = 4 π r2

• The volume of Sphere = 4/3 π r3

• The curved surface area of Cylinder = 2 πrh

• Area of two circular bases = 2 πr2

• The total surface area of Cylinder = Circumference of Cylinder + Curved surface area of Cylinder = 2 πrh + 2 πr2

• Volume of Cylinder = π r2 h

• Slant height of cone = l = √(r2 + h2)

• Curved surface area of cone = πrl

• Total surface area of cone = πr (l + r)

• Volume of cone = ⅓ π r2 h

• Perimeter of cuboid = 4(l + b +h)

• Length of the longest diagonal of a cuboid = √(l2 + b2 + h2)

• Total surface area of cuboid = 2(l×b + b×h + l×h)

• Volume of Cuboid = l × b × h

Students can access extra study materials for Surface Areas and Volume on Vedantu. These resources are available for download, offering additional support for your studies.

• Surface Areas and Volumes Important Questions

• Surface Areas and Volumes Revision Notes

• Surface Areas and Volumes NCERT Exemplar Solutions

• Surface Areas and Volumes RD Sharma Solutions

Chapter 13 - Statistics

This is another excellent chapter in which students learn about the numerical representation of data, whether grouped or ungrouped. Learn how to find the mean, median, and mode of a given dataset. In the following assignment, you will learn about the cumulative frequency distribution and construct cumulative frequency curves.

Topics Covered in Class 10 Statistics

• Mean, median and mode of grouped data. 

• Mean by Direct Method and Assumed Mean Method only.

Important Formulas

There are three ways to determine the mean of the grouped data.

The direct method is: $\bar{x} = \frac{\sum_{i=1}^n f_i x_i}{\sum_{i=1}^n f_i}$ 

Where ∑fi xi is the total of observations from value i = 1 to n. ∑fi represents the number of observations from i = 1 to n.

Assumed Mean Method: $\bar{x} = a+\frac{\sum_{i=1}^n f_i d_i}{\sum_{i=1}^n f_i}$  

The step deviation approach is: $\bar{x} = a+\frac{\sum_{i=1}^n f_i u_i}{\sum_{i=1}^n f_i} \times h$  

Mode of grouped data:

Mode = $l+\frac{f_1-f_0}{2 f_1-f_0-f_2} \times h$ 

The median for grouped data is:

Median = $l+\frac{\frac{n}{2}-c f}{f} \times h$

Students can access extra study materials for Statistics on Vedantu. These resources are available for download, offering additional support for your studies.

• Statistics Important Questions

• Statistics Revision Notes

• Statistics RD Sharma Solutions

Chapter 14 - Probability

Chapter 14 begins with the theoretical approach to probability.  discusses and illustrates the distinction between experimental and theoretical probability.  The chapter is full of examples that help learners understand the idea of probability. Probability is also a very essential chapter for students to focus on, as it will be studied further grades. To stay up with the formulae and new subjects introduced in higher classes, students must first master the fundamentals of the chapter in class 10.

Topics Covered in Class 10 Probability

• Classical definition of probability. 

• Simple problems on finding the probability of an event.

Important Formulas:

The theoretical probability (also known as classical probability) of an event E, denoted as P(E), is

$P(E)=\frac{\text { Number of outcomes favourable to } E}{\text { Number of all possible outcomes of the experiment }}$

We assume that the experiment's outcomes are equally likely.

The probability of a definite event (or certain event) is one.

The chance of an impossible event is zero.

The probability of an occurrence E is a number P(E), where 0 < P(E) ≤ 1.

An elementary occurrence is one that has only one possible conclusion. The total of the probability of all elementary occurrences in an experiment equals one.

Students can access extra study materials for Probability on Vedantu. These resources are available for download, offering additional support for your studies.

• Probability Important Questions

• Probability Revision Notes

• Probability RD Sharma Solutions

CBSE Class 10 Maths Chapter-Wise Marks Weightage

Internal Assessment for CBSE Class 10 Maths

CBSE Class 10 Maths Exam Pattern for 2024-25

The CBSE Class 10 Maths Exam Pattern for 2024-25 consists of two parts:

• Internal Assessment: This is conducted by the school throughout the year and carries 20 marks. It typically includes periodic tests, project work, and other activities.

• Board Exam: This is a 3-hour pen-and-paper exam conducted by the CBSE and carries 80 marks. The exam pattern is as follows:

CBSE Class 10th New Plan 2024-25

The following are some major changes that are planned to be introduced in the new CBSE Exam Pattern:

• Question papers will be designed to assess students' problem-solving and analytical skills.

• The paper design will be revised to include more short answer-type questions, like those ranging from 1 to 5 points.

• Some vocational course exams might be held in February, and final board exams will be completed in March within 15 days.

• Following the early completion of boards, the results are likely to be announced earlier than the existing structure's schedule.

Benefits of Refering to Vedantu’s NCERT Solutions for Class 10 Maths

The Vedantu’s Class 10 NCERT Solutions of Maths provided here in PDFs offer various benefits, including:

• The answers provided here are simple and straightforward.

• To facilitate comprehension, solutions are presented in phases.

• Students are provided diagrams to assist them visualise the solutions.

• All of the questions from each chapter are answered.

• Students are urged to prepare all of the chapters covered in the solution modules, as this will help them obtain a deeper understanding of subjects. 

• For effective preparations, comprehend all of the processes outlined in the answers.

Important Tips for Scoring Well in the Board Exams

Class 10 is an important class in most students’ school lives and is the first class where the students have to appear for a board exam. The class may seem daunting but if the students have been thorough with their studies or the whole academic year then it will not be difficult for them to score good marks at all. To achieve a good score it is necessary that the students keep the following tips and suggestions in mind for the whole year and especially while preparing for the exams.

1. Regular Revision is Key:

• Consistency in revising and practicing daily lessons is crucial.

• Regular practice ensures a better understanding and prevents topics from becoming overwhelming during exams.

2. Practice Extensively:

• Practice all examples and exercises in the NCERT book.

• Utilize solutions PDF for self-assessment and focus on areas with incorrect answers.

• Refer to Vedantu's solutions for additional practice and clarity.

3. Concept Clarity Matters:

• Seek teacher assistance for any doubts promptly.

• Utilize Vedantu's Solutions PDF for easy-to-understand explanations.

• Ensure a clear understanding of all concepts well in advance of exams.

4. Understand Syllabus and Paper Pattern:

• Familiarize yourself with the syllabus, topic weights, and exam pattern.

• This knowledge aids in efficient exam preparation.

5. Effective Time Management:

• Create a timetable considering topic weights and personal strengths/weaknesses.

• Set daily and weekly goals for focused and motivated study sessions.

• Include breaks for effective learning and retention.

6. Identify Strengths and Weaknesses:

• Allocate equal time to all topics; practice weaker areas more.

• Ensure confidence in every topic by thorough practice.

7. Note-taking for Consolidation:

• Write down theorems and formulas for consolidation and quick revision.

• Notes act as a reference point and aid in better memorization.

8. Class 10 Maths Resources:

• Master with our detailed class 10th Maths Guide.

• Practice sample papers and previous years' question papers.

• Understand the question patterns to prepare effectively.

• Vedantu's website offers sample papers for practice.

9. Prioritize Rest and Preparation:

• Ensure a good night's sleep and a healthy meal before the exam.

• Gather necessary materials the night before to avoid last-minute stress.

10. Exam Simulation for Time Management:

• Practice answering sample papers under timed conditions.

• Assess and improve time management skills to attempt all questions.

11. Presenting Answers Effectively:

• Answer questions in the order presented in the paper.

• Highlight final answers and relevant formulas for clarity.

• Avoid leaving any questions unattempted; partial answers can secure marks.

Following the above-mentioned tips and being sincere in your preparation will help you score good marks in the exams. Having a strong base in Maths in Class 10 is definitely beneficial for classes ahead and in the future. This will help you have an edge over others even in competitive exams. Most competitive exams have maths sections and the questions generally asked pertain to what you may have learnt in school. Having a good grip over the formulas and the topics will definitely help you perform well in these exams as well.

CBSE Class 10 Maths Study Materials