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Class 10 Science Notes Chapter 1 — Chemical Reactions and Equations

Mon, 01 Jan 0001 00:00:00 +0000

Class 10 Science Notes Chapter 1 — Chemical Reactions and Equations#

Signs of a chemical reaction#

A chemical reaction forms new substances. You can tell one has happened from:

a change in state
a change in colour
evolution of a gas
a change in temperature
formation of a precipitate

Example: magnesium ribbon burns with a dazzling white flame and forms white magnesium oxide powder.

$$2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$$

Chemical equations#

Reactants are written on the left, products on the right, with an arrow between them.
Balanced equation — the number of atoms of each element is the same on both sides. This follows the law of conservation of mass: mass is neither created nor destroyed in a reaction.
Balance by changing the coefficients in front of formulae, never the subscripts inside them.
States: (s) solid, (l) liquid, (g) gas, (aq) aqueous solution. Conditions such as heat, pressure or a catalyst are written above or below the arrow.

$$3\text{Fe(s)} + 4\text{H}_2\text{O(g)} \rightarrow \text{Fe}_3\text{O}_4\text{(s)} + 4\text{H}_2\text{(g)}$$

Types of chemical reactions#

Type	Pattern	Example
Combination	$A + B \rightarrow AB$	$\text{CaO} + \text{H}_2\text{O} \rightarrow \text{Ca(OH)}_2$
Decomposition	$AB \rightarrow A + B$	$\text{CaCO}_3 \xrightarrow{\text{heat}} \text{CaO} + \text{CO}_2$
Displacement	$A + BC \rightarrow AC + B$	$\text{Fe} + \text{CuSO}_4 \rightarrow \text{FeSO}_4 + \text{Cu}$
Double displacement	$AB + CD \rightarrow AD + CB$	$\text{Na}_2\text{SO}_4 + \text{BaCl}_2 \rightarrow \text{BaSO}_4 + 2\text{NaCl}$

Combination reactions#

Quick lime + water: $\text{CaO} + \text{H}_2\text{O} \rightarrow \text{Ca(OH)}_2$ + heat. Calcium oxide (quick lime) forms calcium hydroxide (slaked lime). A solution of slaked lime is used for whitewashing: it reacts slowly with CO₂ in air to form a shiny coat of calcium carbonate.
Burning of coal: $\text{C} + \text{O}_2 \rightarrow \text{CO}_2$
Formation of water: $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$

Decomposition reactions#

They need energy, given as heat, light or electricity.

Class 10 Maths Notes Chapter 1 — Real Numbers

Mon, 01 Jan 0001 00:00:00 +0000

Class 10 Maths Notes Chapter 1 — Real Numbers#

Key terms#

Prime number — a natural number greater than 1 whose only factors are 1 and itself (2, 3, 5, 7, …).
Composite number — a natural number greater than 1 with a factor other than 1 and itself (4, 6, 9, …).
1 is neither prime nor composite.
HCF — the largest number that divides each of the given numbers exactly.
LCM — the smallest number that is a multiple of each of the given numbers.
Coprime numbers — two numbers whose HCF is 1 (for example, 8 and 15).
Rational number — can be written as $\frac{p}{q}$, where $p$, $q$ are integers and $q \neq 0$.
Irrational number — cannot be written as $\frac{p}{q}$. Examples: $\sqrt{2}$, $\sqrt{3}$, $\sqrt{15}$, $\pi$, $0.10110111\ldots$
Rational and irrational numbers together make up the real numbers.

Fundamental Theorem of Arithmetic#

Theorem 1.1: Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.1

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.1#

Welcome to our comprehensive NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.1. This exercise introduces fundamental concepts of real numbers including prime factorization, Least Common Multiple (LCM), Highest Common Factor (HCF), and composite numbers. Our expert teachers have prepared detailed, step-by-step solutions to help CBSE Class 10 students understand every concept clearly and score full marks in their board exams.

NCERT Solutions for Class 10 Maths Chapter 10 Circles Exercise 10.1

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 10 Maths Chapter 10 Circles Exercise 10.1#

Exercise 10.1 introduces the concept of a tangent to a circle and the number of tangents from a point to a circle. It establishes the fundamental property that a tangent is perpendicular to the radius at the point of tangency.

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

NCERT Solutions for Class 10 Maths Chapter 11 Areas Related to Circles Exercise 11.1

Mon, 01 Jan 0001 00:00:00 +0000

Exercise 11.1 covers areas of sectors, segments, and combinations of plane figures involving circles.

Key Formulas:

Length of arc $= \dfrac{\theta}{360°} \times 2\pi r$
Area of sector $= \dfrac{\theta}{360°} \times \pi r^2$
Area of minor segment $=$ Area of sector $-$ Area of triangle
Use $\pi = \dfrac{22}{7}$ unless stated otherwise.

1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°. #

Answer

Using Area of sector $= \dfrac{\theta}{360°} \times \pi r^2$, we find the area:

NCERT Solutions for Class 10 Maths Chapter 12 Surface Areas and Volumes Exercise 12.1

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 10 Maths Chapter 12 Surface Areas and Volumes Exercise 12.1#

Exercise 12.1 covers the surface area of combinations of solids — cubes, cuboids, cylinders, cones, hemispheres. The total surface area excludes the joined faces.

Key Formulas:

Cylinder: Curved SA = $2\pi rh$, Total SA = $2\pi r(r+h)$
Cone: Curved SA = $\pi rl$, Total SA = $\pi r(r+l)$, where $l = \sqrt{r^2+h^2}$
Sphere: SA = $4\pi r^2$
Hemisphere: Curved SA = $2\pi r^2$, Total SA = $3\pi r^2$
Cube: SA = $6a^2$; Cuboid: SA = $2(lb+bh+hl)$

1. 2 cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid. #

Answer

Let the side of each cube be $a$ cm. The volume of each cube is 64 cm³, so:

NCERT Solutions for Class 10 Maths Chapter 13 Statistics Exercise 13.1

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 10 Maths Chapter 13 Statistics Exercise 13.1#

Exercise 13.1 covers three methods of finding the mean of grouped data: Direct Method, Assumed Mean Method, and Step Deviation Method.

Methods to Find Mean of Grouped Data:

Direct Method: $\bar{x} = \dfrac{\sum f_i x_i}{\sum f_i}$
Assumed Mean Method: $\bar{x} = a + \dfrac{\sum f_i d_i}{\sum f_i}$, where $d_i = x_i - a$
Step Deviation Method: $\bar{x} = a + \dfrac{\sum f_i u_i}{\sum f_i} \times h$, where $u_i = \dfrac{x_i - a}{h}$

1. A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house. #

Answer

Number of plants	Number of houses ($f_i$)	$x_i$	$f_i x_i$
0-2	1	1	1
2-4	2	3	6
4-6	1	5	5
6-8	5	7	35
8-10	6	9	54
10-12	2	11	22
12-14	3	13	39
Total	20		162

Using $\bar{x} = \dfrac{\sum f_i x_i}{\sum f_i}$:

NCERT Solutions for Class 10 Maths Chapter 14 Probability Exercise 14.1

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 10 Maths Chapter 14 Probability Exercise 14.1#

Exercise 14.1 covers classical probability — finding the probability of events using coins, dice, cards, and balls in bags.

Classical Probability:
$$P(E) = \frac{\text{Number of outcomes favourable to } E}{\text{Total number of equally likely outcomes}}$$

$0 \leq P(E) \leq 1$
$P(E) + P(\bar{E}) = 1$
A standard deck has 52 cards: 4 suits (Hearts, Diamonds, Clubs, Spades), 13 each.

1. Complete the following statements: #

(i) Probability of an event E + Probability of the event ’not E’ = ___.

$P(E) + P(\bar{E}) = \mathbf{1}$

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.1

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.1#

Welcome to our comprehensive NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.1. This exercise deals with the geometrical meaning of the zeroes of a polynomial. Students learn to read the number of zeroes directly from the graph of a polynomial by counting the number of times the curve crosses or touches the x-axis. These solutions help CBSE Class 10 students build visual understanding of polynomial behaviour.

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.1

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.1#

Welcome to our comprehensive NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.1. This exercise covers forming and solving pairs of linear equations graphically, checking consistency of pairs of equations, and understanding the geometric interpretation of solutions.

Consistency of a pair of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.1

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.1#

Welcome to our NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.1. This introductory exercise helps students identify quadratic equations and represent real-life situations in the form of quadratic equations. A thorough understanding of this exercise forms the foundation for all subsequent methods of solving quadratic equations.

Quadratic Equation: An equation of the form $ax^2 + bx + c = 0$, where $a \neq 0$ and $a$, $b$, $c$ are real numbers, is called a quadratic equation. The degree of the equation must be exactly 2. If the degree is 1 or 0, it is not quadratic.

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.1

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.1#

Welcome to our NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.1. This introductory exercise helps students identify whether a given sequence is an Arithmetic Progression (AP) and to find the common difference. A solid grasp of these basics is essential for mastering the rest of the chapter.

Arithmetic Progression (AP): A sequence $a_1, a_2, a_3, \ldots$ is an AP if the difference between consecutive terms is constant. This constant is called the Common Difference $d = a_{n+1} - a_n$.

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.1

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.1#

This exercise introduces the concept of similar figures and the criteria for similarity of triangles. Students learn to distinguish between congruent and similar figures, understand that similar figures have the same shape but not necessarily the same size, and apply the AA, SAS, and SSS similarity criteria.

1. Fill in the blanks using the correct word given in brackets: #

(i) All circles are ___. (congruent, similar)

All circles are similar.

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.1

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.1#

Exercise 7.1 focuses on the Distance Formula — finding the distance between two points given their coordinates, and using it to determine properties of geometric figures (collinearity, type of triangle/quadrilateral).

Distance Formula: The distance between two points $P(x_1, y_1)$ and $Q(x_2, y_2)$ is:
$$PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
Distance from Origin: Distance of point $P(x, y)$ from origin $= \sqrt{x^2 + y^2}$

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.1

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.1#

Exercise 8.1 introduces trigonometric ratios — sine, cosine, tangent, cosecant, secant, and cotangent — defined with respect to the sides of a right-angled triangle. Students learn to find these ratios from given information.

Trigonometric Ratios in right △ABC (right angle at B), for the acute angle A:

Ratio Definition Reciprocal

$\sin A$ $\dfrac{\text{side opposite to } A}{\text{hypotenuse}} = \dfrac{BC}{AC}$ $\cosec A = \dfrac{1}{\sin A}$

$\cos A$ $\dfrac{\text{side adjacent to } A}{\text{hypotenuse}} = \dfrac{AB}{AC}$ $\sec A = \dfrac{1}{\cos A}$

$\tan A$ $\dfrac{\text{side opposite to } A}{\text{side adjacent to } A} = \dfrac{BC}{AB}$ $\cot A = \dfrac{1}{\tan A}$

Ratio	Definition	Reciprocal
$\sin A$	$\dfrac{\text{side opposite to } A}{\text{hypotenuse}} = \dfrac{BC}{AC}$	$\cosec A = \dfrac{1}{\sin A}$
$\cos A$	$\dfrac{\text{side adjacent to } A}{\text{hypotenuse}} = \dfrac{AB}{AC}$	$\sec A = \dfrac{1}{\cos A}$
$\tan A$	$\dfrac{\text{side opposite to } A}{\text{side adjacent to } A} = \dfrac{BC}{AB}$	$\cot A = \dfrac{1}{\tan A}$

1. In △ABC, right angled at B, AB = 24 cm, BC = 7 cm. Determine: #

Setup

By Pythagoras Theorem:

NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Exercise 9.1

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Exercise 9.1#

Exercise 9.1 applies trigonometry to real-life problems involving heights and distances. Students use angles of elevation and depression to find unknown heights and distances.

Key Concepts:

Angle of Elevation: The angle formed between the horizontal line and the line of sight when looking upward at an object.
Angle of Depression: The angle formed between the horizontal line and the line of sight when looking downward at an object.
Standard values: $\tan 30° = \frac{1}{\sqrt{3}}$, $\tan 45° = 1$, $\tan 60° = \sqrt{3}$

1. A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30°. #

Answer

Let AB be the pole and AC = 20 m be the rope, with ∠ACB = 30°.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 1 Orienting Yourself Exercise Set 1.1

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 1 — Exercise Set 1.1#

In this exercise you read positions of objects in Reiaan’s room from the Cartesian plane. A point $(x, y)$ lies $x$ units from the y-axis and $y$ units from the x-axis; the origin $O = (0, 0)$ is the bottom-left corner of the room and $1$ unit $= 1$ ft.

Fig. 1.3 shows Reiaan’s room with points OABC marking its corners. The x- and y-axes are marked in the figure. Point O is the origin. Referring to Fig. 1.3, answer the following questions: #

From the figure the relevant readings are: $O = (0, 0)$, $A = (12, 0)$, $B = (12, 10)$, $C = (0, 10)$, room door end-points $D_1 = (8, 0)$ and $R_1 = (11.5, 0)$, bathroom door end-points $B_1 = (0, 1.5)$ and $B_2 = (0, 4)$.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 10 How Quantities Combine Exercise Set 10.1

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 10 — Exercise Set 10.1#

Weighted mean: if values $x_1, x_2, \ldots$ come with weights (sizes) $w_1, w_2, \ldots$,
$$\text{Weighted mean} = \dfrac{w_1x_1 + w_2x_2 + \cdots}{w_1 + w_2 + \cdots}$$
To combine averages of groups of different sizes, weight each average by the size of its group.

1. The average score of students on a test in Section A is 72 and that of students in Section B is 76. What is the combined average of both the sections given that Section A has 30 students and Section B has 25 students? #

Solution

Total marks in Section A $= 72 \times 30 = 2160$, and in Section B $= 76 \times 25 = 1900$.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 11 The World of Algorithms Exercise Set 11.1

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 11 — Exercise Set 11.1#

The book’s algorithm to add two numbers:

Write the numbers one below the other in the Indian base-ten place-value system so that the digits are aligned from right to left.
Add the rightmost digits. If the sum is less than 10, write it below and set carry to 0; otherwise write the units digit below and set carry to 1.
Move to the next column on the left. Add the two digits and the current carry. (Same rule for writing the sum and setting the carry.)
Repeat Step 3 until there are no more digits on the left.
If the carry is 1, write 1 to the left of the bottom row.

1. Add two 4-digit numbers using the steps we have written down. Make sure you follow the steps precisely; do not perform any action that is not explicitly mentioned. Are you able to obtain the correct result? #

Solution

Take $4738 + 5964$.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 12 Quadrilaterals Exercise Set 12.1

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 12 — Exercise Set 12.1#

Definition (from the book): for four distinct points $A, B, C, D$ in a plane, with no three collinear, the points on segments $AB$, $BC$, $CD$, $DA$ form quadrilateral ABCD if every such point other than $A, B, C, D$ lies on exactly one of these segments. A quadrilateral is convex when all its internal angles are less than 180° — equivalently, when its diagonals intersect.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 13 Two Variables, One Line Exercise Set 13.1

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 13 — Exercise Set 13.1#

A linear equation in two variables in standard form is
$$ax + by + c = 0$$
where $a$, $b$, $c$ are real numbers and $a$, $b$ are not both zero. $a$ is the coefficient of $x$, $b$ the coefficient of $y$, and $c$ the constant term.

1. Write a linear equation in two variables in which a = 3, b = 0 and c = −1/5. #

Solution

Substituting in $ax + by + c = 0$:

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 14 Math of Space Exercise Set 14.1

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 14 — Exercise Set 14.1#

Cube of side $a$: surface area $6a^2$, volume $a^3$.

Cuboid $l \times w \times h$: surface area $2(lw + wh + hl)$, volume $lwh$.

1. The volume of a cube is 64 cm³. What is its total surface area? #

Solution

$$a^3 = 64$$

$$\Rightarrow a = 4 \text{ cm}$$$$\text{TSA} = 6a^2$$

$$\text{TSA} = 6 \times 16$$

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 2 Introduction to Linear Polynomials Exercise Set 2.1

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 2 — Exercise Set 2.1#

The degree of a polynomial in one variable is the highest power of that variable that appears with a non-zero coefficient. A non-zero constant has degree $0$, and the polynomials of degree $1, 2, 3$ are called linear, quadratic, cubic, respectively.

1. Find the degrees of the following polynomials: #

(i) 2x² − 5x + 3

Highest power of $x$ is $2$.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 3 The World of Numbers Exercise Set 3.1

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 3 — Exercise Set 3.1#

Natural numbers $\mathbb{N} = \{1, 2, 3, \ldots\}$ are the counting numbers. This exercise uses them for exchange, patterns and counting, and checks whether they are closed under an operation (whether the answer is always a natural number again).

1. A merchant in the port city of Lothal is exchanging bags of spices for copper ingots. He receives 15 ingots for every 2 bags of spices. If he brings 12 bags of spices to the market, how many copper ingots will he leave with? #

Solution

Every 2 bags give 15 ingots, so we count how many groups of 2 bags he has:

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities Exercise Set 4.1

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 4 — Exercise Set 4.1#

$$(a + b)^2 = a^2 + 2ab + b^2$$

To use the identity, decide what $a$ and $b$ are, then write $a^2$, $2ab$ and $b^2$ one after another.

1. Using the identity $(a + b)^2 = a^2 + 2ab + b^2$, expand the following: #

(i) $(7x + 4y)^2$

Here $a = 7x$ and $b = 4y$. We expand $(7x + 4y)^2$:

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I'm Up and Down, and Round and Round Exercise Set 5.1

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 5 — Exercise Set 5.1#

Theorem 1: There is a unique circle passing through three non-collinear points.

Its centre, the circumcentre, is where the perpendicular bisectors of the sides meet, because every point on the perpendicular bisector of $AB$ is equally far from $A$ and $B$.

To draw the circumcircle of a triangle: draw the perpendicular bisectors of any two sides; they meet at the circumcentre $O$. With centre $O$ and radius $OA$, draw the circle. It passes through all three vertices.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 6 Measuring Space Perimeter and Area Exercise Set 6.1

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 6 — Exercise Set 6.1#

Unless stated otherwise, use $\pi \approx \dfrac{22}{7}$.

$$\text{Circumference} = 2\pi r = \pi d$$

$$\text{Arc length} = \dfrac{\theta}{360°} \times 2\pi r$$
Perimeter is the length of the whole boundary — add every curved and straight part, but leave out lines that are inside the shape.

1. The perimeter of a circle is 44 cm. What is its radius? #

Solution

The perimeter (circumference) is $2\pi r$, so:

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 7 The Mathematics of Maybe Exercise Set 7.1

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 7 — Exercise Set 7.1#

Probability is measured on a scale from 0 (impossible) to 1 (certain):

Label	Probability
Impossible	0
Less likely	between 0 and $\dfrac{1}{2}$
Equally likely (even chance)	$\dfrac{1}{2}$
More likely	between $\dfrac{1}{2}$ and 1
Certain	1

1. Rank the following events on a scale from 0 (Impossible) to 1 (Certain). Label each event: Impossible, less likely, equally likely (even chance), more likely, certain. Give reasons why you gave each event its ranking. #

(i) The next Monday will come after Sunday.

Certain — probability 1. Monday always follows Sunday; the order of the days of the week never changes.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 8 Predicting What Comes Next Exercise Set 8.1

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 8 — Exercise Set 8.1#

An explicit rule gives the $n^{th}$ term directly from $n$ (for example $t_n = 5n - 3$). A recursive rule gives each term from the terms before it (for example $t_{n+1} = t_n + 3$), together with the starting term(s).

To check whether a number is a term: set $t_n$ equal to it and solve for $n$. It is a term only if $n$ comes out as a natural number.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 9 Propositions and their Converses Exercise Set 9.1

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 9 — Exercise Set 9.1#

The converse of “if X then Y” is “if Y then X”.

To show a statement is true, give a reason (a proof) that works in every case. To show it is false, one counterexample is enough.

Questions 1–12: Frame the converse for each of the propositions. Then, determine if each of the two statements is true or not. Justify the true statements and give a counterexample for each false statement.

NCERT Class 10 Maths Book

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Class 10 Maths Book#

Tap a chapter to open the official NCERT PDF.

Mathematics#

Front pages (contents, preface) PDF
1 Real Numbers PDF
2 Polynomials PDF
3 Pair of Linear Equations in Two Variables PDF
4 Quadratic Equations PDF
5 Arithmetic Progressions PDF
6 Triangles PDF
7 Coordinate Geometry PDF
8 Introduction to Trigonometry PDF
9 Some Applications of Trigonometry PDF
10 Circles PDF
11 Areas Related to Circles PDF
12 Surface Areas and Volumes PDF
13 Statistics PDF
14 Probability PDF
Appendix 1 · Proofs in Mathematics PDF
Appendix 2 · Mathematical Modelling PDF
Answers PDF

NCERT Class 9 Maths Books

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Class 9 Maths Books#

Tap a chapter to open the official NCERT PDF.

Ganita Manjari Part I#

Front pages (contents, preface) PDF
1 Orienting Yourself: The Use of Coordinates PDF
2 Introduction to Linear Polynomials PDF
3 The World of Numbers PDF
4 Exploring Algebraic Identities PDF
5 I’m Up and Down, and Round and Round PDF
6 Measuring Space: Perimeter and Area PDF
7 The Mathematics of Maybe: Introduction to Probability PDF
8 Predicting What Comes Next: Exploring Sequences and Progressions PDF

Ganita Manjari Part II#

Front pages (contents, preface) PDF
9 Propositions and their Converses PDF
10 How Quantities Combine: Understanding Data PDF
11 The World of Algorithms PDF
12 Quadrilaterals PDF
13 Two Variables, One Line PDF
14 Math of Space: Surface Area and Volume PDF

Class 10 Science Notes Chapter 2 — Acids, Bases and Salts

Mon, 01 Jan 0001 00:00:00 +0000

Class 10 Science Notes Chapter 2 — Acids, Bases and Salts#

Indicators#

Indicator	In acid	In base
Blue litmus	Red	No change
Red litmus	No change	Blue
Methyl orange	Red	Yellow
Phenolphthalein	Colourless	Pink
Turmeric	No change	Reddish-brown

Litmus is a natural indicator, extracted from lichen. Other natural indicators: turmeric, red cabbage, coloured petals (hydrangea, petunia, geranium).
Olfactory indicators change smell. Onion and vanilla lose their smell in a base. Clove oil also loses its smell in a base.
A curry stain (turmeric) on cloth turns reddish-brown with soap (a base), and turns yellow again when rinsed with plenty of water.

Reactions of acids and bases#

Reaction	General equation
Acid + metal	Salt + hydrogen gas
Base + metal (Zn, Al)	Salt + hydrogen gas
Acid + metal carbonate / hydrogencarbonate	Salt + CO₂ + water
Acid + base (neutralisation)	Salt + water
Acid + metal oxide	Salt + water
Base + non-metal oxide	Salt + water

$$\text{Zn} + \text{H}_2\text{SO}_4 \rightarrow \text{ZnSO}_4 + \text{H}_2$$

Class 10 Maths Notes Chapter 2 — Polynomials

Mon, 01 Jan 0001 00:00:00 +0000

Class 10 Maths Notes Chapter 2 — Polynomials#

Key terms#

Polynomial in $x$ — an expression $a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0$, where the powers are whole numbers and the coefficients are real numbers.
Degree — the highest power of $x$ in $p(x)$.
Zero of $p(x)$ — a real number $k$ such that $p(k) = 0$.

Type	Degree	General form
Linear	1	$ax + b$, $a \neq 0$
Quadratic	2	$ax^2 + bx + c$, $a \neq 0$
Cubic	3	$ax^3 + bx^2 + cx + d$, $a \neq 0$

The zero of the linear polynomial $ax + b$ is $-\frac{b}{a}$.
A polynomial of degree $n$ has at most $n$ zeroes. A quadratic has at most 2, a cubic at most 3.

Geometrical meaning of zeroes#

The zeroes of $p(x)$ are exactly the x-coordinates of the points where the graph of $y = p(x)$ meets the x-axis.

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.2

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.2#

Welcome to our NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.2. This exercise asks students to prove that numbers such as $\sqrt{5}$, $3 + 2\sqrt{5}$, $\dfrac{1}{\sqrt{2}}$, $7\sqrt{5}$ and $6 + \sqrt{2}$ are irrational, using the method of proof by contradiction.

Theorem 1.2: Let $p$ be a prime number. If $p$ divides $a^2$, then $p$ divides $a$, where $a$ is a positive integer.

NCERT Solutions for Class 10 Maths Chapter 10 Circles Exercise 10.2

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 10 Maths Chapter 10 Circles Exercise 10.2#

Exercise 10.2 explores properties of tangents drawn from an external point. The key theorem states that tangent segments from an external point to a circle are equal in length.

Theorems used:

The tangent at any point of a circle is perpendicular to the radius through the point of contact.
The lengths of tangents drawn from an external point to a circle are equal. If PA and PB are tangents from an external point P to a circle with centre O, then PA = PB.

1. From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is: (A) 7 cm (B) 12 cm (C) 15 cm (D) 24.5 cm #

Answer

Let O be the centre and QT the tangent, with T the point of contact. Then QT = 24 cm and OQ = 25 cm.

NCERT Solutions for Class 10 Maths Chapter 12 Surface Areas and Volumes Exercise 12.2

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 10 Maths Chapter 12 Surface Areas and Volumes Exercise 12.2#

Exercise 12.2 covers the volume of combinations of solids — finding total volume when solids are combined together.

Key Volume Formulas:

Cylinder: $V = \pi r^2 h$
Cone: $V = \dfrac{1}{3}\pi r^2 h$
Sphere: $V = \dfrac{4}{3}\pi r^3$
Hemisphere: $V = \dfrac{2}{3}\pi r^3$
Cuboid: $V = l \times b \times h$; Cube: $V = a^3$

1. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π. #

Answer

Radius $r$ = 1 cm
Height of cone $h$ = 1 cm

NCERT Solutions for Class 10 Maths Chapter 13 Statistics Exercise 13.2

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 10 Maths Chapter 13 Statistics Exercise 13.2#

Exercise 13.2 covers finding the mode of grouped data using the mode formula.

Mode of Grouped Data:
$$\text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$$
where $l$ = lower class limit of modal class, $f_1$ = frequency of modal class, $f_0$ = frequency of class preceding modal class, $f_2$ = frequency of class succeeding modal class, $h$ = class size.

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.2

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.2#

Welcome to our comprehensive NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.2. This exercise focuses on the relationship between the zeroes and coefficients of a polynomial. Students find the zeroes of quadratic polynomials and verify the relationship, and also form quadratic polynomials when the sum and product of zeroes are given.

Key Relationships for a Quadratic Polynomial $p(x) = ax^2 + bx + c$ with zeroes $\alpha$ and $\beta$:

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.2

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.2#

Welcome to our NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2. This exercise teaches the Substitution Method to solve pairs of linear equations algebraically. The substitution method involves expressing one variable in terms of the other and substituting into the second equation to solve the system.

Substitution Method Steps:

From one equation, express one variable in terms of the other (e.g., $x$ in terms of $y$)
Substitute this expression into the second equation
Solve the resulting single-variable equation
Substitute back to find the other variable

1. Solve the following pair of linear equations by the substitution method. #

(i) $x + y = 14$; $x - y = 4$

$$x + y = 14 \quad \cdots (1)$$

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.2

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.2#

Welcome to our NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.2. This exercise teaches students to solve quadratic equations by the factorisation method. Factorisation involves splitting the middle term and expressing the quadratic as a product of two linear factors, then solving each factor independently.

Factorisation Method: To solve $ax^2 + bx + c = 0$, find two numbers whose sum is $b$ and product is $ac$. Split the middle term, group the terms, factorise, and solve each factor = 0.

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.2

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.2#

Welcome to our NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.2. This exercise develops students’ ability to find the $n$th term of an AP using the formula, to find which term a given number is, and to solve problems using the $n$th term formula. These solutions are essential for CBSE Class 10 board exam preparation.

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.2

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.2#

Exercise 6.2 covers the Basic Proportionality Theorem (also known as Thales’ Theorem) and its converse. The BPT states that if a line is drawn parallel to one side of a triangle, it divides the other two sides in the same ratio.

Basic Proportionality Theorem (BPT / Thales’ Theorem): If a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally. Conversely, if a line divides two sides of a triangle in the same ratio, the line is parallel to the third side.

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.2

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.2#

Exercise 7.2 covers the Section Formula — finding the coordinates of a point that divides a line segment in a given ratio, and its special case (midpoint formula).

Section Formula: The coordinates of point $P$ which divides the line segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ in ratio $m:n$ internally are:
$$P = \left(\frac{mx_2 + nx_1}{m+n},\ \frac{my_2 + ny_1}{m+n}\right)$$
Midpoint Formula: Midpoint of AB $= \left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.2

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.2#

Exercise 8.2 deals with trigonometric ratios of specific angles: 0°, 30°, 45°, 60°, and 90°. Students use these standard values to evaluate expressions.

Standard Trigonometric Values:

Angle	$0°$	$30°$	$45°$	$60°$	$90°$
$\sin$	$0$	$\tfrac{1}{2}$	$\tfrac{1}{\sqrt{2}}$	$\tfrac{\sqrt{3}}{2}$	$1$
$\cos$	$1$	$\tfrac{\sqrt{3}}{2}$	$\tfrac{1}{\sqrt{2}}$	$\tfrac{1}{2}$	$0$
$\tan$	$0$	$\tfrac{1}{\sqrt{3}}$	$1$	$\sqrt{3}$	Not defined
$\cosec$	Not defined	$2$	$\sqrt{2}$	$\tfrac{2}{\sqrt{3}}$	$1$
$\sec$	$1$	$\tfrac{2}{\sqrt{3}}$	$\sqrt{2}$	$2$	Not defined
$\cot$	Not defined	$\sqrt{3}$	$1$	$\tfrac{1}{\sqrt{3}}$	$0$

1. Evaluate the following: #

(i) sin 60° cos 30° + sin 30° cos 60°

We find $\sin 60° \cos 30° + \sin 30° \cos 60°$:

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 1 Orienting Yourself Exercise Set 1.2

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 1 — Exercise Set 1.2#

In this exercise you extend the plane to include negative coordinates and plot points across all four quadrants. Distances between two axis-parallel points are found by subtracting the coordinates that differ; the four quadrants are numbered anticlockwise from the top-right.

On a graph sheet, mark the x-axis and y-axis and the origin O. Mark points from (−7, 0) to (13, 0) on the x-axis and from (0, −15) to (0, 12) on the y-axis. (Use the scale 1 cm = 1 unit.) Using Fig. 1.5, answer the given questions. #

The readings from Fig. 1.5 used below:

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 10 How Quantities Combine Exercise Set 10.2

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 10 — Exercise Set 10.2#

When parts are combined in a ratio such as $3 : 4 : 5$, the ratio gives the weights. First turn each score into the same scale (a fraction or a percentage), then take the weighted mean.

1. Savitri’s marks in Kashmiri are as follows: 35 out of 50 in internal tests, 44 out of 60 in the project, and 80 out of 100 in the final exam. The annual percentage score is calculated by combining the internals, project, and final exam in the ratio 3 : 4 : 5. Which of the following expression(s) gives her annual score (as a percentage) in Kashmiri? #

$$\text{(i)}\ \dfrac{35 \times 3 + 44 \times 4 + 80 \times 5}{3 + 4 + 5} \qquad \text{(ii)}\ \dfrac{\frac{35}{100} \times 3 + \frac{44}{100} \times 4 + \frac{80}{100} \times 5}{3 + 4 + 5}$$

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 11 The World of Algorithms Exercise Set 11.2

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 11 — Exercise Set 11.2#

An algorithm is written as numbered steps, using names for the values we keep track of (such as list-of-divisors) and simple actions we already know how to do (such as checking whether one number divides another).

1. Suppose List 1 and List 2 are two lists of numbers in increasing order. #

(i) Write an algorithm to find elements in List 1 that are not present in List 2.

Algorithm only-in-1

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 12 Quadrilaterals Exercise Set 12.2

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 12 — Exercise Set 12.2#

Theorem 1: In a parallelogram, (a) opposite sides are equal, (b) opposite angles are equal, (c) the diagonals bisect each other.

Theorems 2–5 (tests for a parallelogram): a quadrilateral is a parallelogram if its opposite sides are equal, or its opposite angles are equal, or its diagonals bisect each other, or one pair of opposite sides is both equal and parallel.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 13 Two Variables, One Line Exercise Set 13.2

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 13 — Exercise Set 13.2#

A solution of a linear equation in two variables is an ordered pair $(x, y)$ that makes the equation true. Every linear equation in two variables has infinitely many solutions, and they are exactly the points of its graph — a straight line.

To find a solution: choose any value of $x$, substitute it, and solve for $y$.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 14 Math of Space Exercise Set 14.2

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 14 — Exercise Set 14.2#

Cylinder of radius $r$ and height $h$:
$$\text{CSA} = 2\pi rh, \qquad \text{Volume} = \pi r^2 h$$

1. Two cylinders, A and B, are given. The radius of cylinder B is twice that of cylinder A, and the height of cylinder B is half that of cylinder A. Find the ratio of the curved surface area of A to the curved surface area of B. Also find the ratio of the volume of A to the volume of B. #

Solution

Let A have radius $r$ and height $h$. Then B has radius $2r$ and height $\dfrac{h}{2}$.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 2 Introduction to Linear Polynomials Exercise Set 2.2

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 2 — Exercise Set 2.2#

Evaluating a polynomial means substituting a specific value for the variable and simplifying. Word problems in this exercise translate everyday statements into a linear equation of the form $ax + b = c$, which is then solved for the unknown.

1. Find the value of the linear polynomial 5x − 3 if: #

Let $p(x) = 5x - 3$.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 3 The World of Numbers Exercise Set 3.2

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 3 — Exercise Set 3.2#

Integers $\mathbb{Z} = \{\ldots, -3, -2, -1, 0, 1, 2, 3, \ldots\}$ add zero and the negative numbers to the natural numbers. Brahmagupta described positive numbers as fortunes (dhana) and negative numbers as debts (ṛiṇa), and gave the rules for working with them:

Operation	Rule
Fortune × fortune, or debt × debt	fortune (positive)
Fortune × debt, or debt × fortune	debt (negative)
Subtracting a debt	the same as adding a fortune

1. The temperature in the high-altitude desert of Ladakh is recorded as 4 °C at noon. By midnight, it drops by 15 °C. What is the midnight temperature? #

Solution

A drop of 15 °C means we subtract 15 from the noon temperature:

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities Exercise Set 4.2

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 4 — Exercise Set 4.2#

$$a^2 + 2ab + b^2 = (a + b)^2$$

$$a^2 - 2ab + b^2 = (a - b)^2$$

To factor a perfect square: find $a$ and $b$ whose squares are the first and last terms, then check that the middle term is $2ab$. If the coefficients are awkward, first take out a common factor, as in Example 7.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I'm Up and Down, and Round and Round Exercise Set 5.2

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 5 — Exercise Set 5.2#

Theorem 2: Equal chords of a circle subtend equal angles at the centre.

Theorem 3: Chords of a circle that subtend equal angles at the centre are equal.

All radii of a circle are equal — this fact is used in almost every proof in this chapter.

1. Show that the triangle formed by a chord and the centre of the circle is isosceles. #

Solution

Let $AB$ be a chord of a circle with centre $C$ and radius $r$. Joining $C$ to $A$ and $B$ gives $\triangle CAB$.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 6 Measuring Space Perimeter and Area Exercise Set 6.2

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 6 — Exercise Set 6.2#

$$\text{Area of a triangle} = \dfrac{1}{2} \times \text{base} \times \text{height}$$
Heron’s formula: for sides $a$, $b$, $c$ and semi-perimeter $s = \dfrac{a + b + c}{2}$,
$$\text{Area} = \sqrt{s(s - a)(s - b)(s - c)}$$
Key idea: triangles with the same base and equal heights (their third vertices on a line parallel to the base) have equal areas.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 7 The Mathematics of Maybe Exercise Set 7.2

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 7 — Exercise Set 7.2#

$$\text{Experimental probability} = \dfrac{\text{Number of times the event occurred}}{\text{Total number of trials}}$$

$$\text{Theoretical probability} = \dfrac{\text{Number of favourable outcomes}}{\text{Number of possible outcomes}}$$

To estimate a total from a sample, multiply the sample probability by the size of the whole group.

1. A teacher mixes a large bag of sweets of different colours and randomly selects a sample of 30 sweets. She counts the number of sweets of each colour: 10 red sweets | 8 green sweets | 7 yellow sweets | 5 blue sweets #

(i) Calculate the probability that a randomly picked sweet from the sample is green.

$$P(\text{green}) = \dfrac{8}{30}$$

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 8 Predicting What Comes Next Exercise Set 8.2

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 8 — Exercise Set 8.2#

For an arithmetic progression (AP) with first term $a$ and common difference $d$:
$$t_n = a + (n - 1)d$$

$$\text{Recursive rule: } t_1 = a, \quad t_{n+1} = t_n + d$$

$$S_n = \dfrac{n}{2}\bigl(2a + (n - 1)d\bigr) = \dfrac{n}{2}(\text{first term} + \text{last term})$$

1. Find the 10th and 26th terms of the AP: 3, 8, 13, 18, …. #

Solution

Here $a = 3$ and $d = 8 - 3 = 5$.

NCERT Class 10 Science Book

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Class 10 Science Book#

Tap a chapter to open the official NCERT PDF.

Science#

Front pages (contents, preface) PDF
1 Chemical Reactions and Equations PDF
2 Acids, Bases and Salts PDF
3 Metals and Non-metals PDF
4 Carbon and its Compounds PDF
5 Life Processes PDF
6 Control and Coordination PDF
7 How do Organisms Reproduce? PDF
8 Heredity PDF
9 Light – Reflection and Refraction PDF
10 The Human Eye and the Colourful World PDF
11 Electricity PDF
12 Magnetic Effects of Electric Current PDF
13 Our Environment PDF
Answers PDF

NCERT Class 9 Science Book — Exploration

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Class 9 Science Book#

Tap a chapter to open the official NCERT PDF.

Exploration#

Front pages (contents, preface) PDF
1 Exploration: Entering the World of Secondary Science PDF
2 Cell: The Building Block of Life PDF
3 Tissues in Action PDF
4 Describing Motion Around Us PDF
5 Exploring Mixtures and their Separation PDF
6 How Forces Affect Motion PDF
7 Work, Energy, and Simple Machines PDF
8 Journey Inside the Atom PDF
9 Atomic Foundations of Matter PDF
10 Sound Waves: Characteristics and Applications PDF
11 Reproduction: How Life Continues PDF
12 Patterns in Life: Diversity and Classification PDF
13 Earth as a System: Energy, Matter, and Life PDF

Class 10 Science Notes Chapter 3 — Metals and Non-metals

Mon, 01 Jan 0001 00:00:00 +0000

Class 10 Science Notes Chapter 3 — Metals and Non-metals#

Physical properties#

Property	Metals	Non-metals
Lustre	Shiny (metallic lustre)	Dull
Hardness	Generally hard	Generally soft
Malleable (beaten into sheets)	Yes — gold and silver most	No (brittle)
Ductile (drawn into wires)	Yes — gold most	No
Conduct heat and electricity	Good — silver and copper best	Poor
Sonorous (ringing sound)	Yes (school bells)	No
State at room temperature	Solid (except mercury)	Solid, liquid (bromine) or gas

Exceptions:

Class 10 Maths Notes Chapter 3 — Pair of Linear Equations in Two Variables

Mon, 01 Jan 0001 00:00:00 +0000

Class 10 Maths Notes Chapter 3 — Pair of Linear Equations in Two Variables#

Key terms#

Linear equation in two variables — $ax + by + c = 0$, where $a$, $b$, $c$ are real numbers and $a$, $b$ are not both zero. Its graph is a straight line, and it has infinitely many solutions.
Pair of linear equations — two such equations in the same variables: $$a_1x + b_1y + c_1 = 0$$ $$a_2x + b_2y + c_2 = 0$$
Solution of the pair — a pair of values $(x, y)$ that satisfies both equations.
Consistent — has at least one solution. Inconsistent — has no solution.
Dependent — the two equations represent the same line (infinitely many solutions).

Graphical method and consistency#

Draw both lines on the same axes. Their position tells you the number of solutions, and so do the ratios of the coefficients:

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 1 Orienting Yourself End-of-Chapter Exercises

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 1 — End-of-Chapter Exercises#

These problems use everything from the chapter: quadrants, distances between points, midpoints and shapes on the Cartesian plane. Two results are used again and again:

Distance between $(x_1, y_1)$ and $(x_2, y_2)$:
$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
Midpoint of the segment joining them:
$$M = \left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)$$

1. What are the x-coordinate and y-coordinate of the point of intersection of the two axes? #

Solution

The two axes meet at the origin $O$. The origin is 0 units from both axes.

NCERT Solutions for Class 10 Maths Chapter 13 Statistics Exercise 13.3

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 10 Maths Chapter 13 Statistics Exercise 13.3#

Exercise 13.3 covers finding the median of grouped data using the median formula.

Median of Grouped Data:
$$\text{Median} = l + \frac{\frac{n}{2} - cf}{f} \times h$$
where $l$ = lower class limit of median class, $n$ = total frequency, $cf$ = cumulative frequency of class preceding median class, $f$ = frequency of median class, $h$ = class size. The median class is the class where cumulative frequency first exceeds $\dfrac{n}{2}$.

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.3

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.3#

Welcome to our NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.3. This exercise focuses on the Elimination Method (also called the addition/subtraction method) for solving pairs of linear equations. The method involves making the coefficients of one variable equal and then adding or subtracting the equations to eliminate that variable.

Elimination Method Steps:

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.3

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.3#

Exercise 4.3 covers the nature of roots of a quadratic equation using the discriminant, and word problems that lead to quadratic equations.

Discriminant $D = b^2 - 4ac$ for $ax^2 + bx + c = 0$:

$D > 0$: Two distinct real roots
$D = 0$: Two equal real roots
$D < 0$: No real roots (roots are imaginary)

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

(i) $2x^2 - 3x + 5 = 0$

Here $a = 2,\ b = -3,\ c = 5$.

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.3

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.3#

Welcome to our NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.3. This is the most extensive exercise of the chapter, covering the sum of the first $n$ terms of an AP and its applications in diverse word problems. Mastery of this exercise is crucial for CBSE Class 10 board examination success.

Sum of $n$ Terms of an AP:

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.3

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.3#

Exercise 6.3 deals with the criteria for similarity of triangles — AA (Angle-Angle), SAS (Side-Angle-Side), and SSS (Side-Side-Side). Students apply these criteria to prove triangles similar and find unknown sides and angles.

Similarity Criteria: (1) AA: If two angles of one triangle are equal to two angles of another triangle, the triangles are similar. (2) SAS: If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional, the triangles are similar. (3) SSS: If corresponding sides of two triangles are proportional, the triangles are similar.

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.3

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.3#

Exercise 8.3 covers Trigonometric Identities. Students express trigonometric ratios in terms of other ratios and evaluate expressions using fundamental identities.

Fundamental Identities:

$\sin^2 A + \cos^2 A = 1$
$1 + \tan^2 A = \sec^2 A$
$1 + \cot^2 A = \cosec^2 A$

1. Express the trigonometric ratios $\sin A$, $\sec A$ and $\tan A$ in terms of $\cot A$. #

Answer

sin A in terms of cot A:

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 10 How Quantities Combine Exercise Set 10.3

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 10 — Exercise Set 10.3#

$$\text{Weighted mean} = \dfrac{\sum (\text{weight} \times \text{value})}{\sum \text{weights}}$$
A combined average always lies between the smallest and largest of the values being combined, and is closer to the value with the larger weight.

1. A stationery shop owner made ₹8000 selling books, of which 30% is the profit amount, and ₹1000 selling book covers, of which 50% is the profit amount. What is the percentage of profit on the total sales? #

Solution

$$\text{Profit on books} = 30\% \text{ of } ₹8000 = ₹2400$$

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 11 The World of Algorithms Exercise Set 11.3

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 11 — Exercise Set 11.3#

Recall the two versions of the gcd algorithm in this section:

Original: list the divisors of $m$ and of $n$ in increasing order, list their common divisors, and report the rightmost one.
Last version: start with most-recent-common-divisor = 1, scan $k$ from 2 up to $\min(m, n)$, and update it whenever $k$ divides both numbers.

1. How would our original algorithm change if we computed the divisors of n by examining the numbers from 1 to n in reverse order, from n down to 1? #

Solution

Checking from $n$ down to 1, the divisors are added to the list in decreasing order, for example divisors(18) = [18, 9, 6, 3, 2, 1].

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 12 Quadrilaterals Exercise Set 12.3

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 12 — Exercise Set 12.3#

Theorem 6 (Midpoint Theorem): the segment joining the midpoints of two sides of a triangle is parallel to the third side and has half its length.

Theorem 7 (Converse): the line through the midpoint of one side, parallel to another side, bisects the third side.

Theorem 9 (Varignon): the midpoints of the four sides of a quadrilateral are the vertices of a parallelogram.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 13 Two Variables, One Line Exercise Set 13.3

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 13 — Exercise Set 13.3#

$$\text{Slope} = \dfrac{\text{rise}}{\text{run}} = \dfrac{y_2 - y_1}{x_2 - x_1}$$
A larger slope means a steeper line. A horizontal line has slope 0; a vertical line has no defined slope.

1. The following diagrams represent ski hills. Rank the hills in order of their steepness, from least to greatest. (A: rise 60 m, run 70 m; B: rise 60 m, run 110 m; C: rise 80 m, run 100 m) #

Solution

$$\text{Slope of A} = \dfrac{60}{70} \approx 0.86$$

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 14 Math of Space Exercise Set 14.3

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 14 — Exercise Set 14.3#

Cone of radius $r$, height $h$ and slant height $l$:
$$l^2 = h^2 + r^2$$

$$\text{CSA} = \pi rl, \qquad \text{TSA} = \pi r(l + r), \qquad V = \dfrac{1}{3}\pi r^2 h$$

For $\pi$, use $\dfrac{22}{7}$ or $3.14$.

1. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m. #

Solution

$r = 12$ m and $l = 21$ m.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 2 Introduction to Linear Polynomials Exercise Set 2.3

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 2 — Exercise Set 2.3#

A quantity that changes by the same fixed amount each step can be written as a linear expression $a n + b$, where $b$ is the starting value and $a$ is the constant change per step (positive for growth, negative for decay).

1. A student has ₹500 in her savings bank account. She gets ₹150 every month as pocket money. How much money will she have at the end of every month from the second month onwards? Find a linear expression to represent the amount she will have in the nth month. #

Solution

She starts with ₹500 and adds ₹150 every month, so at the end of the $n^{\text{th}}$ month she has:

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 3 The World of Numbers Exercise Set 3.3

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 3 — Exercise Set 3.3#

A rational number is a number of the form $\dfrac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. To add or subtract rational numbers, first make the denominators equal using the LCM. To multiply, multiply the numerators and the denominators. To divide, multiply by the reciprocal.

Two rational numbers $\dfrac{a}{b}$ and $\dfrac{c}{d}$ are equal when $ad = bc$ (cross-multiplication).

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities Exercise Set 4.3

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 4 — Exercise Set 4.3#

$$(a + b)^2 = a^2 + 2ab + b^2$$

$$(a - b)^2 = a^2 - 2ab + b^2$$

$$(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$$

1. Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier. #

(i) $117^2$

117 splits into three easy parts, $100 + 10 + 7$, so we use the $(a + b + c)^2$ identity. We find $117^2$:

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I'm Up and Down, and Round and Round Exercise Set 5.3

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 5 — Exercise Set 5.3#

Theorem 4: The line joining the centre of a circle and the midpoint of a chord is perpendicular to the chord.

1. Can you explain why the converse to Theorem 4 is true, i.e., why does the perpendicular from the centre of a circle to a chord of the circle bisect the chord? (Hint: Use Fig. 5.12. You are told that ∠CMA = ∠CMB = 90°. You need to show that AM = BM.) #

Solution

Given: a circle with centre $C$, a chord $AB$, and $CM \perp AB$, so $\angle CMA = \angle CMB = 90°$.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 6 Measuring Space Perimeter and Area Exercise Set 6.3

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 6 — Exercise Set 6.3#

Area of a circle: $\pi r^2$

Area of a sector with angle $\theta$ at the centre: $\dfrac{\theta}{360°} \times \pi r^2$

Area of a segment = area of the sector − area of the triangle formed by the chord and the two radii.

Use $\pi = \dfrac{22}{7}$ unless the question says otherwise.

1. Find the area of a sector of a circle with radius 7 cm if the angle of the sector is 60°. #

Solution

$$\text{Area} = \dfrac{60}{360} \times \dfrac{22}{7} \times 7^2$$

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 7 The Mathematics of Maybe Exercise Set 7.3

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 7 — Exercise Set 7.3#

The sample space $S$ lists every possible outcome exactly once; its number of elements $n(S)$ is the sample size. An event is a subset of the sample space.

1. When a single 6-sided die is rolled, what is the total number of possible outcomes in the sample space? #

Solution

$$S = \{1, 2, 3, 4, 5, 6\}$$

∴ There are 6 possible outcomes: $n(S) = 6$.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 8 Predicting What Comes Next Exercise Set 8.3

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 8 — Exercise Set 8.3#

For a geometric progression (GP) with first term $a$ and common ratio $r$:
$$t_n = a\,r^{\,n-1}$$

$$\text{Recursive rule: } t_1 = a, \quad t_{n+1} = r \times t_n$$

1. Find the 12th term of a GP with common ratio 2, whose 8th term is 192. #

Solution

From the 8th term to the 12th term we multiply by $r$ four times:

NCERT Class 10 Social Science Books

Mon, 01 Jan 0001 00:00:00 +0000

Tap a chapter to open the official NCERT PDF.

India and the Contemporary World II (History)#

Front pages (contents, preface) PDF
1 The Rise of Nationalism in Europe PDF
2 Nationalism in India PDF
3 The Making of a Global World PDF
4 The Age of Industrialisation PDF
5 Print Culture and the Modern World PDF

Contemporary India II (Geography)#

Democratic Politics II (Political Science)#

Understanding Economic Development (Economics)#

Front pages (contents, preface) PDF
1 Development PDF
2 Sectors of the Indian Economy PDF
3 Money and Credit PDF
4 Globalisation and the Indian Economy PDF
5 Consumer Rights PDF

NCERT Class 9 Social Science Book — Understanding Society: India and Beyond, Part I

Mon, 01 Jan 0001 00:00:00 +0000

Tap a chapter to open the official NCERT PDF.

Understanding Society: India and Beyond, Part I#

Front pages (contents, preface) PDF
1 Understanding Social Science PDF
2 Shaping of the Earth’s Surface PDF
3 Atmosphere and Climate PDF
4 Early Humans and Beginning of Civilisation PDF
5 State and Society up to 1000 CE PDF
6 Democracy PDF
7 Elections PDF
8 Building Blocks in Economics: The Problem of Choice PDF
9 The Price Puzzle: What Drives the Market PDF

Class 10 Science Notes Chapter 4 — Carbon and its Compounds

Mon, 01 Jan 0001 00:00:00 +0000

Class 10 Science Notes Chapter 4 — Carbon and its Compounds#

Covalent bonding#

Carbon has atomic number 6 (2, 4): 4 valence electrons. It can neither lose 4 nor gain 4 electrons easily, so it shares electrons.
Covalent bond — a bond formed by sharing a pair of electrons. A single bond shares 1 pair, a double bond 2 pairs, a triple bond 3 pairs.
Examples: H₂ (H–H, single), O₂ (O=O, double), N₂ (N≡N, triple), CH₄ (four C–H single bonds).
Covalent compounds have low melting and boiling points (weak forces between molecules) and are poor conductors of electricity (no ions).

Allotropes of carbon#

Allotrope	Structure	Properties
Diamond	Each C bonded to 4 others in a rigid 3-D network	Hardest natural substance; does not conduct
Graphite	Each C bonded to 3 others in hexagonal layers; layers slide	Smooth, slippery; good conductor
Fullerene (C-60)	Carbon atoms in the shape of a football	Named after Buckminster Fuller

Why carbon forms so many compounds#

Catenation — carbon bonds with other carbon atoms to form long chains, branches and rings.
Tetravalency — with a valency of 4, it bonds with other carbon atoms and with H, O, N, S, Cl and others.

The C–C bond is strong, so these compounds are stable.

Class 10 Maths Notes Chapter 4 — Quadratic Equations

Mon, 01 Jan 0001 00:00:00 +0000

Class 10 Maths Notes Chapter 4 — Quadratic Equations#

Key terms#

Quadratic equation — an equation $ax^2 + bx + c = 0$, where $a$, $b$, $c$ are real numbers and $a \neq 0$. This is its standard form.
Root — a real number $\alpha$ is a root if $a\alpha^2 + b\alpha + c = 0$.
The roots of $ax^2 + bx + c = 0$ are the same as the zeroes of the polynomial $ax^2 + bx + c$.
A quadratic equation has at most two roots.
To check whether an equation is quadratic, expand and simplify it first. It is quadratic only if the $x^2$ term remains with a non-zero coefficient (and no higher power remains).

Solving by factorisation#

Write the equation in standard form $ax^2 + bx + c = 0$.
Split the middle term: find two numbers whose sum is $b$ and whose product is $ac$.
Factorise by grouping into two linear factors.
Put each factor equal to zero and solve.

Example: $2x^2 - 5x + 3 = 0$. We need two numbers with sum −5 and product $2 \times 3 = 6$. They are −2 and −3.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 11 The World of Algorithms End-of-Chapter Exercises

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 11 — End-of-Chapter Exercises#

Improved Euclid (Āryabhaṭa’s division) algorithm for gcd(m, n):

If $m < n$, reverse the numbers and compute gcd(n, m).

If $n = 0$, report the answer as $m$.

Otherwise, reduce the problem to computing gcd(n, m mod n), where m mod n is the remainder when $m$ is divided by $n$.

1. Compute the following using the improved version of Euclid’s algorithm. #

Each row divides the larger number by the smaller; the next row uses the divisor and the remainder.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 6 Measuring Space Perimeter and Area End-of-Chapter Exercises

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 6 — End-of-Chapter Exercises#

Unless stated otherwise, use $\pi \approx \dfrac{22}{7}$.

$$\text{Triangle} = \dfrac{1}{2}bh, \qquad \text{Heron: } \sqrt{s(s - a)(s - b)(s - c)}$$

$$\text{Trapezium} = \dfrac{1}{2}(a + b)h, \qquad \text{Rhombus or kite} = \dfrac{1}{2}d_1d_2$$

$$\text{Circle} = \pi r^2, \qquad \text{Sector} = \dfrac{\theta}{360°}\pi r^2$$

1. Identities in algebra can sometimes be shown as area relationships. For example, Fig. 6.41 corresponds to the identity (a + b)² = a² + 2ab + b². Do you see how? Draw figures corresponding to the identities (a + b)(a − b) = a² − b² and (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca. #

How Fig. 6.41 shows (a + b)²

The big square has side $a + b$, so its area is $(a + b)^2$. It is cut into a square $a^2$, a square $b^2$ and two rectangles $ab$. Adding the pieces gives $a^2 + 2ab + b^2$, so $(a + b)^2 = a^2 + 2ab + b^2$.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 8 Predicting What Comes Next End-of-Chapter Exercises

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 8 — End-of-Chapter Exercises#

AP: $t_n = a + (n - 1)d$, $\quad S_n = \dfrac{n}{2}\bigl(2a + (n - 1)d\bigr)$

GP: $t_n = a\,r^{\,n-1}$

Sum of the first $n$ natural numbers: $\dfrac{n(n + 1)}{2}$

1. Find the 31st term of an AP whose 11th term is 38 and 16th term is 73. #

Solution

The 11th term is 38, so:

NCERT Class 10 English Books

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Class 10 English Books#

Tap a chapter to open the official NCERT PDF.

First Flight#

Front pages (contents, preface) PDF
1 A Letter to God PDF
2 Nelson Mandela: Long Walk to Freedom PDF
3 Two Stories about Flying PDF
4 From the Diary of Anne Frank PDF
5 Glimpses of India PDF
6 Mijbil the Otter PDF
7 Madam Rides the Bus PDF
8 The Sermon at Benares PDF
9 The Proposal PDF

Footprints without Feet#

Words and Expressions 2 (Workbook)#

Front pages (contents, preface) PDF
1 A Letter to God PDF
2 Nelson Mandela: Long Walk to Freedom PDF
3 Two Stories about Flying PDF
5 Glimpses of India PDF
6 Mijbil the Otter PDF
7 Madam Rides the Bus PDF
8 The Sermon at Benares PDF

NCERT Class 9 English Book — Kaveri

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Class 9 English Book#

Tap a chapter to open the official NCERT PDF.

Kaveri#

Front pages (contents, preface) PDF
1 How I Taught My Grandmother to Read PDF
2 The Pot Maker PDF
3 Winds of Change PDF
4 Vitamin-M PDF
5 The World of Limitless Possibilities PDF
6 Twin Melodies PDF
7 Carrier of Words PDF
8 Follow That Dream PDF
Appendix PDF

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.4

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.4#

Welcome to our NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.4 (Optional Exercise). This advanced exercise tests deeper understanding of APs through challenging applications and proofs. These solutions equip CBSE Class 10 students with higher-order problem solving skills for competitive examinations and board assessments.

1. Which term of the AP: $121, 117, 113, \ldots$ is its first negative term? [Hint: Find $n$ for $a_n < 0$] #

Answer

Here $a = 121$ and:

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 10 How Quantities Combine Exercise Set 10.4

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 10 — Exercise Set 10.4#

In a stacked bar (or column) chart, the full length of a bar shows a total, and the coloured pieces show the parts that make it up. A piece’s size is read as the difference between its top and bottom ends.

1. The following stacked column chart shows the number of animal species in the IUCN Red List, by class, over time. #

(i) What does the number 14,234 on the top of the column 2019 represent?

It is the total number of animal species on the IUCN Red List in 2019 — the sum of all the classes (mammals, reptiles, birds, insects, amphibians, molluscs, fish and others) stacked in that column.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 12 Quadrilaterals Exercise Set 12.4

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 12 — Exercise Set 12.4#

In any tiling, the angles of the tiles that meet around a point must add up to exactly 360° — no gaps and no overlaps.

1. Justify why the plane cannot be tiled with a regular pentagon. #

Solution

Each angle of a regular pentagon is:

$$\dfrac{(5 - 2) \times 180°}{5} = 108°$$

Look at a corner of a tile. Around that point the angles must total 360°.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 13 Two Variables, One Line Exercise Set 13.4

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 13 — Exercise Set 13.4#

A pair of linear equations in two variables is written as

$$a_1x + b_1y + c_1 = 0, \qquad a_2x + b_2y + c_2 = 0$$

Each fact in the situation gives one equation.

Given below are some everyday situations. Construct a pair of linear equations in two variables for each situation.

1. On two different days a family buys movie tickets and snack boxes. Let the cost of one movie ticket be ₹x and the cost of one snack box be ₹y. (i) On the first day, the family buys 2 movie tickets and 3 snack boxes for ₹850. (ii) On the second day, the family buys 4 movie tickets and 1 snack box for ₹1100. #

Solution

On the first day, 2 tickets and 3 snack boxes cost ₹850, so:

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 14 Math of Space Exercise Set 14.4

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 14 — Exercise Set 14.4#

Sphere: $\text{SA} = 4\pi r^2$, $V = \dfrac{4}{3}\pi r^3$

Hemisphere: $\text{CSA} = 2\pi r^2$, $\text{TSA} = 3\pi r^2$, $V = \dfrac{2}{3}\pi r^3$

1. A ball bearing has a radius of 0.7 cm. Find its surface area. #

Solution

$$\text{SA} = 4 \times \dfrac{22}{7} \times 0.7^2$$

$$\text{SA} = 4 \times \dfrac{22}{7} \times 0.49$$

$$\text{SA} = 6.16 \text{ cm}^2$$

*2. Two solid spheres made of the same metal have weights 5920 g and 740 g. Determine the radius of the larger sphere, if the diameter of the smaller one is 5 cm. #

Solution

Same metal, so weight is proportional to volume, which is proportional to $r^3$:

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 2 Introduction to Linear Polynomials Exercise Set 2.4

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 2 — Exercise Set 2.4#

A relation $y = a t + b$ is called linear because $y$ changes by a constant amount $a$ for each unit change in $t$. If $a > 0$ we call it linear growth; if $a < 0$ we call it linear decay.

1. Suppose a plant has height 1.75 feet and it grows by 0.5 feet each month. #

The plant starts at 1.75 feet and grows 0.5 feet each month, so after $t$ months its height in feet is:

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 3 The World of Numbers Exercise Set 3.4

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 3 — Exercise Set 3.4#

To place $\dfrac{p}{q}$ on the number line, divide each unit into $q$ equal parts and move $p$ parts from 0 — to the right if the number is positive, to the left if it is negative. Between any two rational numbers there are infinitely many rational numbers; one of them is their average $\dfrac{a + b}{2}$.

1. Represent the rational numbers $\dfrac{2}{3}$, $-\dfrac{5}{4}$ and $1\dfrac{1}{2}$ on a single number line. #

Solution

We first write each number as a fraction and see between which integers it lies:

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities Exercise Set 4.4

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 4 — Exercise Set 4.4#

$$(x + a)(x + b) = x^2 + (a + b)x + ab$$

$$(a + b)(a - b) = a^2 - b^2$$

To factor $x^2 + px + q$: find two numbers whose sum is $p$ and product is $q$, then split the middle term. For $ax^2 + bx + c$, find two numbers whose sum is $b$ and product is $ac$.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I'm Up and Down, and Round and Round Exercise Set 5.4

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 5 — Exercise Set 5.4#

Theorem 6: Chords of a circle having the same length are all at the same distance from the centre.

Theorem 7: Chords of a circle that are equidistant from the centre have equal length.

The distance of a chord from the centre is the length of the perpendicular from the centre to the chord. By Theorem 5, this perpendicular meets the chord at its midpoint.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 7 The Mathematics of Maybe Exercise Set 7.4

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 7 — Exercise Set 7.4#

A tree diagram lists all outcomes of a multi-step experiment: one branch for each outcome of the first step, then branches from each of those for the second step. Every path from start to end is one outcome.

1. There are two fruit baskets A and B. Basket A has one apple and two oranges. Basket B has one banana and one mango. You randomly pick one fruit from each basket. #

(i) Draw a tree diagram showing all possible pairs of fruits.

We label the oranges Orange 1 and Orange 2 so that all fruits in basket A are equally likely to be picked.

Class 10 Maths Notes Chapter 5 — Arithmetic Progressions

Mon, 01 Jan 0001 00:00:00 +0000

Class 10 Maths Notes Chapter 5 — Arithmetic Progressions#

Key terms#

Arithmetic progression (AP) — a list of numbers in which each term, except the first, is got by adding a fixed number $d$ to the term before it.
First term — $a$ (also written $a_1$).
Common difference — $d = a_{k+1} - a_k$. It can be positive, negative or zero.
General form — $a,\ a + d,\ a + 2d,\ a + 3d,\ \ldots$
Finite AP has a last term, written $l$. An infinite AP has no last term.

Each term is the one before it plus the common difference $d$.

To check whether a list is an AP, find $a_2 - a_1$, $a_3 - a_2$, $a_4 - a_3$, … If all the differences are equal, it is an AP. One unequal difference is enough to show it is not.

Class 10 Science Notes Chapter 5 — Life Processes

Mon, 01 Jan 0001 00:00:00 +0000

Class 10 Science Notes Chapter 5 — Life Processes#

Life processes are the basic functions that keep an organism alive: nutrition, respiration, transportation and excretion. Multicellular organisms need specialised tissues for these, since diffusion alone is too slow.

Nutrition#

Type	Meaning	Examples
Autotrophic	Makes its own food from simple inorganic substances (CO₂, water) using sunlight	Green plants, some bacteria
Heterotrophic	Depends on others for food	Animals, fungi

Heterotrophs take food in different ways: breaking it down outside the body and absorbing it (fungi like bread mould, yeast, mushrooms), taking in whole food and digesting it inside (most animals), or living on a host (parasites like cuscuta, ticks, lice, tapeworms).

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 12 Quadrilaterals End-of-Chapter Exercises

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 12 — End-of-Chapter Exercises#

Parallelogram tests: opposite sides equal; opposite angles equal; diagonals bisect each other; one pair of sides equal and parallel.

Midpoint Theorem and its converse (Theorem 7); Centroid Theorem (medians meet at a point dividing each in the ratio 2 : 1); Varignon’s theorem.

1. Using a fact about parallelograms, show how to tile the plane using any given triangle. (Hint: Can you use the parallelogram tiling in the introduction?) #

Solution

Take the triangle $ABC$ and a copy turned through 180° about the midpoint $M$ of side $BC$. The half-turn sends $B$ to $C$ and $C$ to $B$, and sends $A$ to a new point $D$ with $M$ the midpoint of $AD$.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 14 Math of Space End-of-Chapter Exercises

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 14 — End-of-Chapter Exercises#

Guesstimates (marked below) have no single exact answer. Make a guess first, state your assumptions clearly, and then calculate. Different sensible assumptions give different — but reasonable — answers.

1. Estimate how many scoops of ice cream can be obtained from a cuboidal container of dimensions 10 cm × 15 cm × 20 cm. #

Guesstimate

$$V_{\text{container}} = 10 \times 15 \times 20$$

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 7 The Mathematics of Maybe End-of-Chapter Exercises

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 7 — End-of-Chapter Exercises#

$$P(E) = \dfrac{\text{Number of favourable outcomes}}{\text{Number of possible outcomes}}, \qquad 0 \leq P(E) \leq 1$$

$$P(\text{not } E) = 1 - P(E)$$
In a tree diagram, multiply the probabilities along a path, and add the probabilities of different paths.

1. Fill in the blanks. #

Answers

Statement	Blank
(i) The probability of an impossible event is ___.	0
(ii) The set of all possible outcomes of a random experiment is called the ___.	sample space
(iii) The probability of an event that is certain to happen is ___.	1
(iv) Tossing a fair coin has a probability of ___ for getting heads.	$\dfrac{1}{2}$

2. In a survey of 50 students, 15 students said they liked football. The number of students who like football is 15, and the _ (frequency/relative frequency) is _ (fill in the fraction or decimal). #

Solution

The count 15 is the frequency. The fraction of all students is the relative frequency:

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 10 How Quantities Combine Exercise Set 10.5

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 10 — Exercise Set 10.5#

In a 100% stacked bar chart, every bar has the same length (100%), and each piece shows a share of the total. It is good for comparing proportions, but it hides the totals.

1. The smart watch data of 5 people was tracked from Monday to Friday. The recorded data tells us how many hours each person spent lying down, sitting, and standing or moving around, etc. #

	Lying down	Sitting	Standing or moving around
Pavani (Patient)	20	3	1
Raghu (Nurse)	7	4	13
Zakir (Teacher)	8	6	10
Sahana (Student)
Julie (______)	8	10	6

(i) How much time does Sahana spend per day in each of these body-states? Make a reasonable guess and fill her row in the table.

Answers will vary. The three numbers must add up to 24 hours. A school student sleeps about 8–9 hours, sits in class and while studying for many hours, and moves about at school, at play and doing chores. A reasonable guess:

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 13 Two Variables, One Line Exercise Set 13.5

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 13 — Exercise Set 13.5#

For $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:

Ratios Lines Solutions

$\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$ intersect exactly one

$\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$ parallel none

$\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$ coincident infinitely many

Ratios	Lines	Solutions
$\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$	intersect	exactly one
$\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$	parallel	none
$\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$	coincident	infinitely many

1. Form a pair of linear equations for each of the following problems and find their solutions. #

(i) The sum of two integers is +5 and their difference is −21. Find the two numbers.

Let the numbers be $x$ and $y$.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 2 Introduction to Linear Polynomials Exercise Set 2.5

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 2 — Exercise Set 2.5#

To find the constants $a$ and $b$ in $y = a x + b$ from two known points $(x_1, y_1)$ and $(x_2, y_2)$, subtract the two equations to eliminate $b$: $a = \dfrac{y_2 - y_1}{x_2 - x_1}$, then substitute back to get $b$.

1. A learning platform charges a fixed monthly fee and an additional cost per digital learning module accessed. A student observes that when she accessed 10 modules, her bill was ₹400. When she accessed 14 modules, her bill was ₹500. If the monthly bill y depends on the number of modules accessed, x, according to the relation y = ax + b, find the values of a and b. #

Solution

When she accessed 10 modules, her bill was ₹400, so:

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 3 The World of Numbers Exercise Set 3.5

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 3 — Exercise Set 3.5#

The decimal expansion of a rational number either terminates or repeats. The decimal expansion of an irrational number never ends and never repeats.

Let $\dfrac{p}{q}$ be in its lowest terms. Its decimal terminates exactly when the prime factors of $q$ are only 2s and/or 5s. Otherwise the decimal repeats.

1. Without performing long division, determine which of the following rational numbers will have terminating decimals and which will be repeating: $\dfrac{7}{20}$, $\dfrac{4}{15}$ and $\dfrac{13}{250}$. Then check your answers by explicitly performing the long divisions and expressing these rational numbers as decimals. #

Prediction (without division)

Each fraction is already in its lowest terms. We factorise each denominator:

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities Exercise Set 4.5

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 4 — Exercise Set 4.5#

To simplify a rational expression: factor the numerator and the denominator completely, then cancel the factors they share (which is allowed because the denominator is not zero).

Identities used in this exercise:

$$(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$$

$$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$$

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I'm Up and Down, and Round and Round Exercise Set 5.5

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 5 — Exercise Set 5.5#

Theorem 8: If $AB > DE$ are two chords of a circle, then $AB$ is closer to the centre than $DE$. The longer chord is nearer the centre.

For a chord at distance $d$ from the centre of a circle of radius $r$, the perpendicular from the centre, half the chord and a radius form a right triangle.

NCERT Class 10 Hindi Books

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Class 10 Hindi Books#

Tap a chapter to open the official NCERT PDF.

क्षितिज 2 (Kshitij)#

कृतिका 2 (Kritika)#

स्पर्श 2 (Sparsh)#

संचयन 2 (Sanchayan)#

NCERT Class 9 Hindi Book — गंगा (Ganga)

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Class 9 Hindi Book#

Tap a chapter to open the official NCERT PDF.

गंगा (Ganga)#

Class 10 Science Notes Chapter 6 — Control and Coordination

Mon, 01 Jan 0001 00:00:00 +0000

Class 10 Science Notes Chapter 6 — Control and Coordination#

Animals coordinate their activities with the nervous system and hormones. Plants use only hormones.

Nervous tissue#

Receptors — specialised tips of nerve cells that detect information from the environment (in the sense organs). For example, gustatory receptors detect taste and olfactory receptors detect smell.
Neuron (nerve cell) — the unit of the nervous system.

Path of a nerve impulse:
dendrite → cell body → axon → nerve ending → synapse → next neuron or muscle

Class 10 Maths Notes Chapter 6 — Triangles

Mon, 01 Jan 0001 00:00:00 +0000

Class 10 Maths Notes Chapter 6 — Triangles#

Similar figures#

Similar figures have the same shape but not necessarily the same size.
Congruent figures have the same shape and the same size.
All congruent figures are similar, but similar figures need not be congruent.
All circles are similar. All squares are similar. All equilateral triangles are similar.

Two polygons with the same number of sides are similar if both:

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 10 How Quantities Combine End-of-Chapter Exercises

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 10 — End-of-Chapter Exercises#

$$\text{Weighted mean} = \dfrac{\sum (\text{weight} \times \text{value})}{\sum \text{weights}}$$
To combine averages, weight each one by how many items it is the average of.

1. In cricket, the run rate is the average number of runs scored per over. In a T20 match, a team scored 6 runs in the first over making the run rate 6. #

(i) In the second over they scored 12 runs. What is the run rate now?

$$\text{Run rate} = \dfrac{6 + 12}{2}$$

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 13 Two Variables, One Line End-of-Chapter Exercises

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 13 — End-of-Chapter Exercises#

Slope-intercept form: $y = mx + c$, where $m$ is the slope and $c$ is the y-intercept.

For a pair of equations: $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$ → one solution; $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$ → none; $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$ → infinitely many.

1. The graph of the line y = 3x, passing through (0, 0) and (2, 6) is given below. #

(i) Identify the slope of the line from the graph and explain how you calculated it using the two points on the line.

From $(0, 0)$ to $(2, 6)$, the line rises 6 while running 2:

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 3 The World of Numbers End-of-Chapter Exercises

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 3 — End-of-Chapter Exercises#

1. Convert the following rational numbers in the form of a terminating decimal or non-terminating and repeating decimal, whichever the case may be, by the process of long division: #

(i) $\dfrac{3}{50}$

Bring down	÷ 50	Remainder
30	0	30
300	6	0

The remainder becomes 0, so the decimal stops:

$$\dfrac{3}{50} = 0.06$$

This is a terminating decimal.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities End-of-Chapter Exercises

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 4 — End-of-Chapter Exercises#

$$(a \pm b)^2 = a^2 \pm 2ab + b^2$$

$$(a + b)(a - b) = a^2 - b^2$$

$$(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$$

$$(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$$

$$(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$$

$$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$$

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 2 Introduction to Linear Polynomials Exercise Set 2.6

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 2 — Exercise Set 2.6#

Every equation of the form $y = a x + b$ has a straight-line graph. The number $a$ is the slope (steepness and direction of the line) and $b$ is the y-intercept (the point where the line crosses the y-axis).

1. Draw the graphs of the following sets of lines. In each case, reflect on the role of ‘a’ and ‘b’. #

(i) y = 4x, y = 2x, y = x

Pick a few $x$-values and find $y$:

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I'm Up and Down, and Round and Round Exercise Set 5.6

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 5 — Exercise Set 5.6#

Theorem 9: The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc at any point on the circle outside the arc.

So angles in the same segment are equal, and the angle in a semicircle is 90°.

1. In a circle with centre O, the central angle AOB is 60°. If the radius of the circle is 12 cm, what is the length of the chord AB? #

Solution

$OA$ and $OB$ are radii, so $OA = OB = 12$ cm and $\triangle OAB$ is isosceles. Its base angles are equal, so:

NCERT Class 10 Sanskrit Books

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Class 10 Sanskrit Books#

Tap a chapter to open the official NCERT PDF.

शेमुषी 2 (Shemushi)#

व्याकरणवीथिः (Vyakaranavithi)#

अभ्यासवान् भव 2 (Abhyaswaan Bhav)#

NCERT Class 9 Sanskrit Books

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Class 9 Sanskrit Books#

Tap a chapter to open the official NCERT PDF.

शारदा (Sharada)#

ऋषिकुल्या (Risikulya, R2)#

Class 10 Maths Notes Chapter 7 — Coordinate Geometry

Mon, 01 Jan 0001 00:00:00 +0000

Class 10 Maths Notes Chapter 7 — Coordinate Geometry#

Distance formula#

The distance between $P(x_1, y_1)$ and $Q(x_2, y_2)$ is
$$PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
The distance of $P(x, y)$ from the origin is
$$OP = \sqrt{x^2 + y^2}$$

The distance $PQ$ is the hypotenuse of a right triangle with legs $x_2 - x_1$ and $y_2 - y_1$, so Pythagoras gives the formula.

Class 10 Science Notes Chapter 7 — How do Organisms Reproduce?

Mon, 01 Jan 0001 00:00:00 +0000

Class 10 Science Notes Chapter 7 — How do Organisms Reproduce?#

Why reproduce#

Reproduction is not needed to keep one individual alive, but it keeps the species going.
The basic event is DNA copying, along with making extra cell apparatus. DNA holds the information for making proteins, and so for the body design.
Copying is never perfect, so it produces small variations. Variations help some individuals survive when the environment changes (for example, heat-resistant bacteria when water gets warmer). Variation is useful for the survival of the species, not necessarily for the individual.

Asexual reproduction#

A new organism comes from a single parent. The offspring are almost identical to the parent.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 2 Introduction to Linear Polynomials End-of-Chapter Exercises

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 2 — End-of-Chapter Exercises#

These problems combine everything from the chapter: writing polynomials of a chosen degree, evaluating polynomials, translating word problems into linear equations, and graphing $y = a x + b$. Key formula used often: for a line through two points $(x_1, y_1)$ and $(x_2, y_2)$, slope $a = \dfrac{y_2 - y_1}{x_2 - x_1}$.

1. Write a polynomial of degree 3 in the variable x, in which the coefficient of the x² term is −7. #

Solution (answers vary)

Any polynomial whose highest power of $x$ is 3 and whose $x^2$ term is $-7x^2$ works. For example:

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I'm Up and Down, and Round and Round End-of-Chapter Exercises

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 5 — End-of-Chapter Exercises#

Results used in this exercise

The perpendicular from the centre to a chord bisects it, and the line from the centre to the midpoint of a chord is perpendicular to it (Theorems 4 and 5).

A chord at distance $d$ from the centre of a circle of radius $r$ has length $2\sqrt{r^2 - d^2}$.

Equal chords are equidistant from the centre, and the longer chord is nearer the centre (Theorems 6–8).

The angle at the centre is double the angle at the circle (Theorem 9); the angle in a semicircle is 90°.

Opposite angles of a cyclic quadrilateral add up to 180°, and the converse holds (Theorems 11 and 12).

1. In a circle, a chord is 5 cm away from the centre. If the radius of the circle is 13 cm, what is the length of the chord? #

Solution

Here $r = 13$ cm and $d = 5$ cm. We find the length of the chord:

NCERT Class 10 Health and Physical Education Book

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Class 10 Health and Physical Education Book#

Tap a chapter to open the official NCERT PDF.

Health and Physical Education#

Front pages (contents, preface) PDF
1 Physical Education: Relation with other Subjects PDF
2 Effects of Physical Activities on Human Body PDF
3 Growth and Development during Adolescence PDF
4 Individual Games and Sports I PDF
5 Individual Games and Sports II PDF
7 Team Games and Sports II PDF
9 Dietary Considerations and Food Quality PDF
10 Safety Measures for Healthy Living PDF
11 Healthy Community Living PDF
12 Social Health PDF

NCERT Class 9 Physical Education Book — Khel Praveen

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Class 9 Physical Education Book#

Tap a chapter to open the official NCERT PDF.

Khel Praveen#

Front pages (contents, preface) PDF
Unit 1 Evolution of Physical Education and Well-being PDF
Unit 2 Science and Sports PDF
Unit 3 Olympism PDF
Unit 4 Sports for All PDF
Unit 5 Sports PDF
Unit 6 Yoga for Holistic Health and Personal Excellence PDF

Class 10 Science Notes Chapter 8 — Heredity

Mon, 01 Jan 0001 00:00:00 +0000

Class 10 Science Notes Chapter 8 — Heredity#

Variation and heredity#

Heredity — the passing of traits from parents to offspring.
Variation — differences between individuals of a species.
Asexual reproduction gives very little variation (only small copying errors in DNA). Sexual reproduction gives much more variation, because it combines DNA from two parents.
Variations build up over generations. Some help survival in a changing environment; these individuals are selected by nature.

Inherited traits#

A child shares basic features with the parents but is not identical to them.
Each trait is influenced by both the father’s and the mother’s DNA, so every child has two versions of each trait.

Mendel’s experiments#

Gregor Johann Mendel (1822–1884) worked with garden peas (Pisum sativum), using contrasting traits: round/wrinkled seeds, tall/short plants, violet/white flowers, yellow/green seeds. He counted the offspring of each type in every generation.

Class 10 Maths Notes Chapter 8 — Introduction to Trigonometry

Mon, 01 Jan 0001 00:00:00 +0000

Class 10 Maths Notes Chapter 8 — Introduction to Trigonometry#

Trigonometric ratios#

In $\triangle ABC$, right-angled at $B$, for the acute angle $A$:

NCERT Class 9 Arts Book — Madhurima

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Class 9 Arts Book#

Tap a chapter to open the official NCERT PDF.

Madhurima#

Front pages (contents, preface) PDF
1 Introduction to the History of Arts PDF
2 The Stage Is Yours PDF
3 Symbols and Metaphors on Stage PDF
4 Creativity in Light and Sound PDF
5 From Vision to Performance: Direction and Acting PDF
6 Music and Us PDF
7 The Science of Music PDF
8 Rāga and Tāla PDF
9 Music Travels Around the World PDF
10 Components of Dance PDF
11 Dance as Visual Storytelling PDF
12 Cultural Connections Through Dance PDF
13 Dance Choreography PDF
14 Exploring the Fundamentals of Visual Arts PDF
15 Mediums and Techniques in Visual Art PDF
16 Visual Composition PDF
17 Branches of Visual Arts PDF

Class 10 Maths Notes Chapter 9 — Some Applications of Trigonometry

Mon, 01 Jan 0001 00:00:00 +0000

Class 10 Maths Notes Chapter 9 — Some Applications of Trigonometry#

Key terms#

Line of sight — the line from the observer’s eye to the point on the object being viewed.
Horizontal — the line through the observer’s eye, parallel to the ground.
Angle of elevation — the angle between the line of sight and the horizontal when the object is above eye level (you raise your head).
Angle of depression — the angle between the line of sight and the horizontal when the object is below eye level (you lower your head).

The angle of depression from A to B equals the angle of elevation from B to A. They are alternate angles, because both horizontals are parallel.

Class 10 Science Notes Chapter 9 — Light: Reflection and Refraction

Mon, 01 Jan 0001 00:00:00 +0000

Class 10 Science Notes Chapter 9 — Light: Reflection and Refraction#

Reflection#

Laws of reflection:

The angle of incidence equals the angle of reflection ($\angle i = \angle r$).

The incident ray, the normal at the point of incidence and the reflected ray all lie in the same plane.

These laws hold for all reflecting surfaces, including curved ones.

Plane mirror image: virtual, erect, the same size as the object, as far behind the mirror as the object is in front, and laterally inverted.

NCERT Class 9 Skill Education Book — Kaushal Vikas

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Class 9 Skill Education Book#

Tap a chapter to open the official NCERT PDF.

Kaushal Vikas#

Front pages (contents, preface) PDF
1 Introduction to Agricultural Practices PDF
2 Rooftop Gardening PDF
3 Precision Farming PDF
4 Additional Vocations (Work with Life Forms) PDF
5 Shaping Materials PDF
6 Construction PDF
7 Apparel PDF
8 Additional Vocations (Work with Machines and Materials) PDF
9 Personal and Lifestyle Services PDF
10 Healthcare PDF
11 Tourism PDF
12 Additional Vocations (Work in Human Services) PDF

Class 10 Maths Notes Chapter 10 — Circles

Mon, 01 Jan 0001 00:00:00 +0000

Class 10 Maths Notes Chapter 10 — Circles#

Key terms#

Secant — a line that meets a circle at two points.
Tangent — a line that meets a circle at exactly one point. That point is the point of contact.
A tangent is the special case of a secant when its two points of intersection come together.
Length of a tangent — the length of the segment from an external point to the point of contact.
A line and a circle can have 0, 1 or 2 common points: non-intersecting, tangent, or secant.
A circle has infinitely many tangents, one at each point. It can have at most two parallel tangents, at the ends of a diameter.

Tangent and radius#

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

Class 10 Science Notes Chapter 10 — The Human Eye and the Colourful World

Mon, 01 Jan 0001 00:00:00 +0000

Class 10 Science Notes Chapter 10 — The Human Eye and the Colourful World#

The human eye#

Part	Function
Cornea	Thin transparent front membrane; most of the refraction happens at its outer surface
Iris	Dark muscular diaphragm that controls the size of the pupil
Pupil	Opening that controls how much light enters (small in bright light, large in dim light)
Eye lens	Fine-adjusts the focal length to focus on the retina; forms a real, inverted image
Ciliary muscles	Change the curvature, and so the focal length, of the eye lens
Retina	Light-sensitive screen with light-sensitive cells; sends electrical signals
Optic nerve	Carries signals from the retina to the brain, which sees the image upright

Power of accommodation — the eye lens’s ability to change its focal length.
- Distant object: ciliary muscles relax, the lens gets thinner, and its focal length increases.
- Near object: ciliary muscles contract, the lens gets thicker, and its focal length decreases.
Near point (least distance of distinct vision) of a normal adult eye: 25 cm. Far point: infinity.
Cataract — the lens becomes milky and cloudy, causing partial or complete loss of vision. It is treated by surgery.

Defects of vision#

Defect	Can see clearly	Image forms	Causes	Corrected with
Myopia (near-sightedness)	Near objects only	In front of the retina	Eyeball too long, or lens too curved	Concave lens
Hypermetropia (far-sightedness)	Distant objects only; near point is beyond 25 cm	Behind the retina	Eyeball too short, or focal length too long	Convex lens
Presbyopia	Not near objects (with age)	Behind the retina	Ciliary muscles weaken and the lens loses flexibility	Convex lens; bifocal lens if both defects

A bifocal lens has a concave upper part (distant vision) and a convex lower part (near vision).
Defects can also be corrected with contact lenses or surgery.
Eye donation: one pair of eyes can give vision to up to four people with corneal blindness. Eyes must be removed within 4–6 hours of death.

Refraction through a prism#

A prism has two triangular bases and three rectangular faces. The angle between two side faces is the angle of the prism.
Light bends towards the base of the prism.
Angle of deviation — the angle between the incident ray (extended) and the emergent ray.

Dispersion#

Dispersion — splitting of white light into its colours by a prism.

Class 10 Maths Notes Chapter 11 — Areas Related to Circles

Mon, 01 Jan 0001 00:00:00 +0000

Key terms#

Sector — the region between an arc and the two radii to its ends. The smaller one is the minor sector, the larger one the major sector.
Segment — the region between a chord and its arc. The smaller one is the minor segment, the larger one the major segment.
Angle of the sector — the angle $\theta$ that the two radii make at the centre.
For a minor sector of angle $\theta$, the major sector has angle $360° - \theta$.

OAB with the arc AB is the minor sector. The region between chord AB and arc AB is the minor segment.

Class 10 Science Notes Chapter 11 — Electricity

Mon, 01 Jan 0001 00:00:00 +0000

Class 10 Science Notes Chapter 11 — Electricity#

Electric current#

$$I = \frac{Q}{t}$$

Electric current — the rate of flow of charge. In metal wires, the charges that move are electrons.
SI unit: ampere (A). 1 A = 1 C/s. Smaller units: 1 mA = 10⁻³ A, 1 μA = 10⁻⁶ A.
SI unit of charge: coulomb (C). Charge of an electron = $1.6 \times 10^{-19}$ C, so 1 C is the charge of about $6 \times 10^{18}$ electrons.
Conventional current flows from positive to negative, opposite to the flow of electrons.
Ammeter measures current. It is connected in series and has low resistance.

Potential difference#

$$V = \frac{W}{Q}$$

Potential difference — work done to move a unit charge from one point to another.
SI unit: volt (V). 1 V = 1 J/C. A cell (or battery) keeps up the potential difference by chemical action.
Voltmeter measures potential difference. It is connected in parallel and has high resistance.

Circuit symbols#

Component	Symbol
Electric cell	─\|‖─ (long line +)
Battery	several cells in series
Plug key / switch (open, closed)	─( )─ , ─(•)─
Wire joint / crossing without joint	dot at the joint / wires crossing
Bulb	⊗ in a circle
Resistor	─///─
Variable resistor (rheostat)	resistor with an arrow
Ammeter / voltmeter	Ⓐ / Ⓥ (circles)

Ohm’s law#

At constant temperature, the current through a metal wire is directly proportional to the potential difference across it:

Class 10 Science Notes Chapter 12 — Magnetic Effects of Electric Current

Mon, 01 Jan 0001 00:00:00 +0000

Class 10 Science Notes Chapter 12 — Magnetic Effects of Electric Current#

Hans Christian Oersted (1820) found that a current in a wire deflects a compass needle. This showed that electricity and magnetism are linked.
A compass needle is a small bar magnet. Its north-seeking end is its north pole.

Magnetic field and field lines#

Magnetic field — the region around a magnet where its force can be felt. It has both size and direction.
Field lines show the field. The direction is the way the north pole of a compass would move.

Properties of magnetic field lines:

Class 10 Maths Notes Chapter 12 — Surface Areas and Volumes

Mon, 01 Jan 0001 00:00:00 +0000

Class 10 Maths Notes Chapter 12 — Surface Areas and Volumes#

Formulas for basic solids#

The measurements used in the formulas: $r$ radius, $h$ height, $l$ slant height (cone).

Solid	Curved / lateral surface area	Total surface area	Volume
Cuboid ($l \times b \times h$)	$2h(l + b)$	$2(lb + bh + hl)$	$lbh$
Cube (edge $a$)	$4a^2$	$6a^2$	$a^3$
Cylinder ($r$, $h$)	$2\pi rh$	$2\pi r(r + h)$	$\pi r^2 h$
Cone ($r$, $h$, $l$)	$\pi rl$	$\pi r(r + l)$	$\frac{1}{3}\pi r^2 h$
Sphere ($r$)	$4\pi r^2$	$4\pi r^2$	$\frac{4}{3}\pi r^3$
Hemisphere ($r$)	$2\pi r^2$	$3\pi r^2$	$\frac{2}{3}\pi r^3$

Slant height of a cone:

Class 10 Science Notes Chapter 13 — Our Environment

Mon, 01 Jan 0001 00:00:00 +0000

Class 10 Science Notes Chapter 13 — Our Environment#

Ecosystem#

Ecosystem — all the living organisms in an area, together with the non-living parts of their environment, interacting with each other.
- Biotic components: plants, animals, microorganisms.
- Abiotic components: temperature, rainfall, wind, soil, minerals.
Natural ecosystems: forests, ponds, lakes. Human-made (artificial) ecosystems: gardens, crop fields, an aquarium.

Component	Role	Examples
Producers	Make food from sunlight by photosynthesis (autotrophs)	Green plants, blue-green algae
Consumers	Eat producers or other consumers	Herbivores, carnivores, omnivores, parasites
Decomposers	Break down dead remains and waste into simple substances, returning nutrients to the soil	Bacteria and fungi

Without decomposers, dead matter would pile up and nutrients would never return to the soil.

Class 10 Maths Notes Chapter 13 — Statistics

Mon, 01 Jan 0001 00:00:00 +0000

Class 10 Maths Notes Chapter 13 — Statistics#

Key terms#

Class mark $x_i$ — the mid-point of a class: $$x_i = \frac{\text{upper limit} + \text{lower limit}}{2}$$
Class size $h$ — upper limit minus lower limit.
Cumulative frequency ($cf$) of a class — the sum of the frequencies of that class and all classes before it.
Modal class — the class with the highest frequency.
Median class — the first class whose cumulative frequency is greater than $\frac{n}{2}$.

Mean of grouped data#

Each class is represented by its class mark. Use any one method:

Class 10 Maths Notes Chapter 14 — Probability

Mon, 01 Jan 0001 00:00:00 +0000

Class 10 Maths Notes Chapter 14 — Probability#

Key terms#

Experiment — an action with well-defined outcomes, such as tossing a coin.
Outcome — a possible result of the experiment.
Equally likely outcomes — each outcome has the same chance of happening. A fair coin, a fair die and a well-shuffled deck give equally likely outcomes.
Event — a collection of outcomes, such as “getting an even number”.
Elementary event — an event with only one outcome.
Sure (certain) event — always happens. Impossible event — never happens.

Theoretical probability#

Class 10 Maths — All Formulas (Quick Revision)

Mon, 01 Jan 0001 00:00:00 +0000

Class 10 Maths — All Formulas#

Every formula in the latest NCERT Class 10 Maths book, chapter by chapter.

1. Real Numbers#

Every composite number = product of primes, in exactly one way (apart from order).
HCF = product of the smallest powers of the common primes.
LCM = product of the greatest powers of all the primes.
For two numbers only: $$\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$$
If a prime $p$ divides $a^2$, then $p$ divides $a$.

2. Polynomials#

For $ax^2 + bx + c$ with zeroes $\alpha$, $\beta$:

Class 10 Science — All Formulas (Quick Revision)

Mon, 01 Jan 0001 00:00:00 +0000

Class 10 Science — All Formulas#

Every formula, rule and key fact in the latest NCERT Class 10 Science book.

Light — reflection and refraction#

Formulas#

Quantity	Formula
Radius of curvature	$R = 2f$
Mirror formula	$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$
Magnification (mirror)	$m = \dfrac{h'}{h} = -\dfrac{v}{u}$
Lens formula	$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$
Magnification (lens)	$m = \dfrac{h'}{h} = \dfrac{v}{u}$
Snell’s law	$\dfrac{\sin i}{\sin r} = \text{constant}$
Absolute refractive index	$n_m = \dfrac{c}{v}$
Relative refractive index	$n_{21} = \dfrac{v_1}{v_2}$
Power of a lens	$P = \dfrac{1}{f}$
Lenses in contact	$P = P_1 + P_2 + P_3 + \cdots$

Notes on the formulas#

Formula	Explanation and notes
Radius of curvature	$R$ = radius of curvature, $f$ = focal length. Holds for spherical mirrors of small aperture.
Mirror formula	$u$ = object distance, $v$ = image distance, $f$ = focal length, all measured from the pole. Has a + sign.
Magnification (mirror)	$h$ = object height, $h'$ = image height. Has a minus sign.
Lens formula	Distances measured from the optical centre. Has a − sign.
Magnification (lens)	No minus sign.
Snell’s law	$i$ = angle of incidence, $r$ = angle of refraction. The constant is the refractive index of medium 2 with respect to medium 1, for a given pair of media and colour of light.
Absolute refractive index	$c$ = speed of light in vacuum ($3 \times 10^8$ m/s), $v$ = speed of light in the medium. No unit.
Relative refractive index	$v_1$, $v_2$ = speed of light in medium 1 and medium 2. Higher $n$ means optically denser.
Power of a lens	$f$ must be in metres. Unit: dioptre (D); 1 D is the power of a lens with $f$ = 1 m.
Lenses in contact	Net power is the sum of the individual powers, with signs.

Laws#

Law	Statement
Reflection 1	$\angle i = \angle r$
Reflection 2	Incident ray, normal and reflected ray lie in the same plane
Refraction 1	Incident ray, refracted ray and normal lie in the same plane
Refraction 2 (Snell’s law)	$\dfrac{\sin i}{\sin r}$ is constant for a given pair of media and colour of light
Rarer → denser	Light bends towards the normal
Denser → rarer	Light bends away from the normal

New Cartesian sign convention#

Quantity	Sign
Object position	Always on the left, so $u$ is always negative
Distances measured from	Pole (mirror) or optical centre (lens)
Along incident light (right)	Positive
Against incident light (left)	Negative
Height above the principal axis	Positive
Height below the principal axis	Negative

Mirror and lens signs#

	Concave mirror	Convex mirror	Convex lens	Concave lens
Type	Converging	Diverging	Converging	Diverging
Focal length $f$	Negative	Positive	Positive	Negative
Power $P$	—	—	Positive	Negative
Real image possible?	Yes	No	Yes	No

Magnification	Meaning
$m$ negative	Real, inverted image
$m$ positive	Virtual, erect image
$\lvert m \rvert > 1$	Enlarged
$\lvert m \rvert < 1$	Diminished

Image formation — concave mirror and convex lens#

Positions in brackets are for the lens (F₁, 2F₁ and O in place of F, C and P).

Padhaiwise

Class 10 Science Notes Chapter 1 — Chemical Reactions and Equations

Class 10 Science Notes Chapter 1 — Chemical Reactions and Equations#

Signs of a chemical reaction#

Chemical equations#

Types of chemical reactions#

Combination reactions#

Decomposition reactions#

Class 10 Maths Notes Chapter 1 — Real Numbers

Class 10 Maths Notes Chapter 1 — Real Numbers#

Key terms#

Fundamental Theorem of Arithmetic#

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.1

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.1#

NCERT Solutions for Class 10 Maths Chapter 10 Circles Exercise 10.1

NCERT Solutions for Class 10 Maths Chapter 10 Circles Exercise 10.1#

NCERT Solutions for Class 10 Maths Chapter 11 Areas Related to Circles Exercise 11.1

NCERT Solutions for Class 10 Maths Chapter 11 Areas Related to Circles Exercise 11.1#

1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°. #

NCERT Solutions for Class 10 Maths Chapter 12 Surface Areas and Volumes Exercise 12.1

NCERT Solutions for Class 10 Maths Chapter 12 Surface Areas and Volumes Exercise 12.1#

1. 2 cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid. #

NCERT Solutions for Class 10 Maths Chapter 13 Statistics Exercise 13.1

NCERT Solutions for Class 10 Maths Chapter 13 Statistics Exercise 13.1#

1. A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house. #

NCERT Solutions for Class 10 Maths Chapter 14 Probability Exercise 14.1

NCERT Solutions for Class 10 Maths Chapter 14 Probability Exercise 14.1#

1. Complete the following statements: #

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.1

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.1#

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.1

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.1#

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.1

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.1#

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.1

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.1#

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.1

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.1#

1. Fill in the blanks using the correct word given in brackets: #

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.1

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.1#

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.1

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.1#

1. In △ABC, right angled at B, AB = 24 cm, BC = 7 cm. Determine: #

NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Exercise 9.1

NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Exercise 9.1#

1. A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30°. #

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 1 Orienting Yourself Exercise Set 1.1

NCERT Solutions for Class 9 Maths Chapter 1 — Exercise Set 1.1#

Fig. 1.3 shows Reiaan’s room with points OABC marking its corners. The x- and y-axes are marked in the figure. Point O is the origin. Referring to Fig. 1.3, answer the following questions: #

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 10 How Quantities Combine Exercise Set 10.1

NCERT Solutions for Class 9 Maths Chapter 10 — Exercise Set 10.1#

1. The average score of students on a test in Section A is 72 and that of students in Section B is 76. What is the combined average of both the sections given that Section A has 30 students and Section B has 25 students? #

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 11 The World of Algorithms Exercise Set 11.1

NCERT Solutions for Class 9 Maths Chapter 11 — Exercise Set 11.1#

1. Add two 4-digit numbers using the steps we have written down. Make sure you follow the steps precisely; do not perform any action that is not explicitly mentioned. Are you able to obtain the correct result? #

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 12 Quadrilaterals Exercise Set 12.1

NCERT Solutions for Class 9 Maths Chapter 12 — Exercise Set 12.1#

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 13 Two Variables, One Line Exercise Set 13.1

NCERT Solutions for Class 9 Maths Chapter 13 — Exercise Set 13.1#

1. Write a linear equation in two variables in which a = 3, b = 0 and c = −1/5. #

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 14 Math of Space Exercise Set 14.1

NCERT Solutions for Class 9 Maths Chapter 14 — Exercise Set 14.1#

1. The volume of a cube is 64 cm³. What is its total surface area? #

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 2 Introduction to Linear Polynomials Exercise Set 2.1

NCERT Solutions for Class 9 Maths Chapter 2 — Exercise Set 2.1#

1. Find the degrees of the following polynomials: #

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 3 The World of Numbers Exercise Set 3.1

NCERT Solutions for Class 9 Maths Chapter 3 — Exercise Set 3.1#

1. A merchant in the port city of Lothal is exchanging bags of spices for copper ingots. He receives 15 ingots for every 2 bags of spices. If he brings 12 bags of spices to the market, how many copper ingots will he leave with? #

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities Exercise Set 4.1

NCERT Solutions for Class 9 Maths Chapter 4 — Exercise Set 4.1#

1. Using the identity $(a + b)^2 = a^2 + 2ab + b^2$, expand the following: #

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I'm Up and Down, and Round and Round Exercise Set 5.1

NCERT Solutions for Class 9 Maths Chapter 5 — Exercise Set 5.1#

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 6 Measuring Space Perimeter and Area Exercise Set 6.1

NCERT Solutions for Class 9 Maths Chapter 6 — Exercise Set 6.1#

1. The perimeter of a circle is 44 cm. What is its radius? #

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 7 The Mathematics of Maybe Exercise Set 7.1

NCERT Solutions for Class 9 Maths Chapter 7 — Exercise Set 7.1#