📐 CBSE Class 10 Mathematics
Complete Revision Notes with Solved Examples | All 14 Chapters
Based on Latest CBSE Syllabus 2024-25
Real Numbers
📌 The Fundamental Theorem of Arithmetic
Every composite number can be expressed as a product of prime numbers and this factorisation is unique, apart from the order in which the prime factors occur. This is also called the Unique Factorisation Theorem.
Key Statement
Any integer greater than 1 can either be a prime number or can be written as a unique product of prime numbers.
Examples:
- \(2 \times 11 = 22\) is the same as \(11 \times 2 = 22\)
- \(6 = 2 \times 3\) or \(3 \times 2\), where 2 and 3 are prime numbers
- \(15 = 3 \times 5\) or \(5 \times 3\), where 3 and 5 are prime numbers
- \(12 = 2 \times 2 \times 3 = 2^2 \times 3\)
📌 HCF and LCM by Prime Factorisation Method
Key Formulae
Step 1: Find all the prime factors of given numbers.
Step 2: \(\text{HCF} = \text{Product of the smallest power of each common prime factor}\)
Step 3: \(\text{LCM} = \text{Product of the greatest power of each prime factor involved}\)
Relationship:
\[ \text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b \]
\[ \text{HCF}(a,b) = \frac{a \times b}{\text{LCM}(a,b)} \quad \text{and} \quad \text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)} \]
Number System Hierarchy: N ⊂ W ⊂ Z ⊂ Q ⊂ R
📌 Rational and Irrational Numbers
Rational Numbers
A number in the form \(\dfrac{p}{q}\), where \(p\) and \(q\) are co-prime numbers and \(q \neq 0\), is known as a rational number.
Examples: \(2, -3, \dfrac{3}{7}, \dfrac{2}{5}, -\dfrac{5}{6}\), etc.
Irrational Numbers
A number that cannot be written in the form \(\dfrac{p}{q}\), where \(p\) and \(q\) are integers and \(q \neq 0\).
Examples: \(\sqrt{2}, \sqrt{3}, \sqrt{5}\), etc.
📌 Proof of Irrationality
Theorem: \(\sqrt{2}\) is irrational
Proof: Assume \(\sqrt{2} = \dfrac{a}{b}\) in its simplest form.
Squaring both sides: \(2b^2 = a^2\), meaning \(a^2\) is even, so \(a\) is even.
Let \(a = 2c\), then \(2b^2 = 4c^2 \Rightarrow b^2 = 2c^2\), so \(b\) is also even.
This contradicts the assumption that \(\dfrac{a}{b}\) is in simplest form. Hence, \(\sqrt{2}\) is irrational.
Theorem: \(\sqrt{3}\) is irrational
Proof: Assume \(\sqrt{3} = \dfrac{a}{b}\). Squaring: \(3b^2 = a^2\), meaning \(a\) is divisible by 3.
If \(a\) is divisible by 3, then \(b\) is also divisible by 3. This contradicts simplest form. Hence, \(\sqrt{3}\) is irrational.
📌 Solved Examples
Show that \(\dfrac{7}{4}\) is a rational number.
Solution: A rational number is any number that can be expressed as \(\dfrac{p}{q}\), where \(p\) and \(q\) are integers and \(q \neq 0\).
Since \(\dfrac{7}{4}\) is in the form \(\dfrac{p}{q}\), with \(p = 7\) and \(q = 4\), and both are integers, \(\dfrac{7}{4}\) is a rational number. \(\blacksquare\)
Prove that \(\sqrt{5}\) is irrational.
Solution: \(\sqrt{5}\) is not a perfect square, meaning it cannot be written as a fraction \(\dfrac{p}{q}\) where \(p\) and \(q\) are integers.
By the method of contradiction (similar to \(\sqrt{2}\)), assume \(\sqrt{5} = \dfrac{a}{b}\) in simplest form.
Squaring: \(5b^2 = a^2\). So 5 divides \(a^2\), which means 5 divides \(a\).
Let \(a = 5c\), then \(5b^2 = 25c^2 \Rightarrow b^2 = 5c^2\), so 5 divides \(b\).
This contradicts that \(\dfrac{a}{b}\) is in simplest form. Hence, \(\sqrt{5}\) is irrational. \(\blacksquare\)
Polynomials
📌 Definition and Types
A polynomial in one variable \(x\) of degree \(n\) is an algebraic expression of the form:
where \(n\) is a whole number and \(a_0, a_1, a_2, \ldots, a_n\) are real numbers.
Based on Number of Terms:
- Monomial: One term — e.g., \(2x, -4x^2, \dfrac{3}{4}x\)
- Binomial: Two terms — e.g., \(2x + 3, 5y^2 - 2y\)
- Trinomial: Three terms — e.g., \(4x^2 - 3x + 4\)
Based on Degree:
- Linear: Degree 1 — e.g., \(2x + 4\)
- Quadratic: Degree 2 — e.g., \(3x^2 - 2x + 4\)
- Cubic: Degree 3 — e.g., \(x^3 - x^2 + x + 1\)
📌 Zeroes of a Polynomial
A real number \(k\) is a zero of polynomial \(p(x)\) if \(p(k) = 0\).
Geometrically, the zeroes are the x-coordinates where the graph of \(y = p(x)\) intersects the x-axis.
Graphs of Linear, Quadratic, and Cubic Polynomials showing zeroes on x-axis
📌 Relationship Between Zeroes and Coefficients
For Quadratic Polynomial \(ax^2 + bx + c\):
\[ \text{Sum of zeroes } (\alpha + \beta) = -\frac{b}{a} \]
\[ \text{Product of zeroes } (\alpha \beta) = \frac{c}{a} \]
| Type | General Form | Max Zeroes | Relationship |
|---|---|---|---|
| Linear | \(ax + b\) | 1 | \(k = -\dfrac{b}{a}\) |
| Quadratic | \(ax^2 + bx + c\) | 2 | \(\alpha + \beta = -\dfrac{b}{a}\), \(\alpha\beta = \dfrac{c}{a}\) |
📌 Discriminant of a Quadratic Polynomial
For \(f(x) = ax^2 + bx + c\), the discriminant \(D = b^2 - 4ac\) determines the nature of zeroes:
- \(D > 0\): Two distinct real zeroes
- \(D = 0\): Two equal (coincident) zeroes
- \(D < 0\): No real zeroes
📌 Solved Examples
Find the zeros of the polynomial: \(f(x) = 2x^2 - 8x + 6\)
Solution: Set \(f(x) = 0\):
\[ 2x^2 - 8x + 6 = 0 \]
Factorise: \(2(x^2 - 4x + 3) = 2(x - 1)(x - 3) = 0\)
Setting each factor to zero:
\(x - 1 = 0 \Rightarrow x = 1\)
\(x - 3 = 0 \Rightarrow x = 3\)
∴ The zeros are \(x = 1\) and \(x = 3\).
Evaluate \(P(x) = x^3 - 4x^2 + 5x - 2\) for \(x = 2\).
Solution: Substitute \(x = 2\):
\[ P(2) = (2)^3 - 4(2)^2 + 5(2) - 2 \]
\[ = 8 - 16 + 10 - 2 = 0 \]
∴ \(P(x) = 0\) when \(x = 2\), confirming \(x = 2\) is a zero.
Pair of Linear Equations in Two Variables
📌 General Form
The most general form of a pair of linear equations in two variables:
where \(a_1^2 + b_1^2 \neq 0\) and \(a_2^2 + b_2^2 \neq 0\).
📌 Conditions for Solutions
Key Conditions
Unique Solution (Intersecting Lines): \(\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}\) → Consistent
No Solution (Parallel Lines): \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}\) → Inconsistent
Infinite Solutions (Coincident Lines): \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}\) → Consistent
Three possible relationships between pairs of linear equations
📌 Algebraic Methods
1. Substitution Method
- Express one variable in terms of the other from one equation
- Substitute into the other equation
- Solve for the remaining variable
- Back-substitute to find the other variable
2. Elimination Method
- Multiply equations to make coefficients equal
- Add or subtract to eliminate one variable
- Solve for the remaining variable
- Back-substitute to find the other
📌 Solved Examples
Solve: \(x + 2y = 11\) and \(2x - y = 1\)
Solution (Substitution Method):
From equation (2): \(y = 2x - 1\)
Substitute in equation (1):
\[ x + 2(2x - 1) = 11 \]
\[ x + 4x - 2 = 11 \Rightarrow 5x = 13 \Rightarrow x = \frac{13}{5} \]
\[ y = 2\left(\frac{13}{5}\right) - 1 = \frac{26}{5} - \frac{5}{5} = \frac{21}{5} \]
∴ Solution: \(\left(\dfrac{13}{5}, \dfrac{21}{5}\right)\)
Find equations of lines parallel and perpendicular to \(y = 2x + 3\) passing through \((1, -2)\).
Solution:
Parallel line: Same slope \(m = 2\)
\[ y - (-2) = 2(x - 1) \Rightarrow y + 2 = 2x - 2 \]
\[ \boxed{y = 2x - 4} \]
Perpendicular line: Slope \(= -\dfrac{1}{2}\)
\[ y + 2 = -\frac{1}{2}(x - 1) \Rightarrow y = -\frac{1}{2}x + \frac{1}{2} - 2 \]
\[ \boxed{y = -\frac{1}{2}x - \frac{3}{2}} \]
🧠 Mnemonic: PIC — NUI
Parallel → No solution | Intersecting → Unique solution | Coincident → Infinite solutions
Quadratic Equations
📌 Standard Form
A quadratic equation in variable \(x\) is of the form:
📌 Methods of Solving
Method 1: Factorisation (Splitting the Middle Term)
- Find product \(ac\)
- Find two numbers whose product is \(ac\) and sum is \(b\)
- Split middle term and factorise by grouping
- Set each factor to zero
Method 2: Quadratic Formula (Sridharacharya's Formula)
Quadratic Formula
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(b^2 - 4ac \geq 0\) for real roots.
📌 Discriminant and Nature of Roots
For \(ax^2 + bx + c = 0\), Discriminant \(D = b^2 - 4ac\):
- \(D > 0\): Two distinct real roots
- \(D = 0\): Two equal real roots
- \(D < 0\): No real roots
📌 Solved Examples
Solve: \(x^2 + 5x + 6 = 0\)
Solution: Factorise:
\[ x^2 + 5x + 6 = (x + 2)(x + 3) = 0 \]
\(x + 2 = 0 \Rightarrow x = -2\)
\(x + 3 = 0 \Rightarrow x = -3\)
∴ Solutions: \(x = -2\) and \(x = -3\)
Solve: \(2x^2 - 7x + 3 = 0\)
Solution: Product \(ac = 2 \times 3 = 6\). Find two numbers: product = 6, sum = \(-7\). Numbers are \(-6\) and \(-1\).
\[ 2x^2 - 6x - x + 3 = 0 \]
\[ 2x(x - 3) - 1(x - 3) = 0 \]
\[ (2x - 1)(x - 3) = 0 \]
\(2x - 1 = 0 \Rightarrow x = \dfrac{1}{2}\)
\(x - 3 = 0 \Rightarrow x = 3\)
∴ Solutions: \(x = \dfrac{1}{2}\) and \(x = 3\)
Arithmetic Progressions
📌 Definition
An Arithmetic Progression (A.P.) is a sequence of numbers in which each term is obtained by adding a fixed number \(d\) (common difference) to the preceding term.
General form: \(a, a+d, a+2d, a+3d, \ldots\)
📌 Key Formulae
nth Term (from beginning):
\[ a_n = a + (n-1)d \]
nth Term (from end):
\[ a_n = l - (n-1)d \quad \text{where } l \text{ is the last term} \]
Sum of n terms:
\[ S_n = \frac{n}{2}[2a + (n-1)d] \]
Or when last term \(l\) is known:
\[ S_n = \frac{n}{2}[a + l] \]
nth term from sum:
\[ a_n = S_n - S_{n-1} \]
🧠 Mnemonic for nth term
"Nokia Offers Additional Programmers In English To Attract Positive New One Buyer Daily"
\(a_n = a + (n-1)d\)
📌 Key Facts
- If a constant is added/subtracted from each term of A.P., the result is also an A.P.
- If each term is multiplied/divided by a non-zero constant, the result is also an A.P.
- If three consecutive numbers \(a, b, c\) are in A.P., then \(2b = a + c\).
- Brahmagupta is known as the father of Arithmetic.
📌 Solved Examples
Which term of the A.P. \(6, 13, 20, 27, \ldots\) is 98 more than its 24th term?
Solution: Here, \(a = 6\), \(d = 13 - 6 = 7\)
According to the question: \(a_n = a_{24} + 98\)
\[ a + (n-1)d = a + (24-1)d + 98 \]
\[ 7(n-1) = 23 \times 7 + 98 \]
\[ n - 1 = 23 + 14 = 37 \Rightarrow n = 38 \]
∴ The 38th term is the required term.
If \(S_n = 3n^2 + 7n\), find the first term and common difference.
Solution:
For \(n = 1\): \(S_1 = a = 3(1)^2 + 7(1) = 10\)
For \(n = 2\): \(S_2 = 3(4) + 7(2) = 12 + 14 = 26\)
Second term \(= S_2 - S_1 = 26 - 10 = 16\)
Common difference \(d = 16 - 10 = 6\)
∴ First term \(a = 10\), Common difference \(d = 6\)
Coordinate Geometry
📌 Basic Concepts
- Cartesian Plane: Formed by two perpendicular axes — X-axis (horizontal) and Y-axis (vertical)
- Origin: Point of intersection \(O(0, 0)\)
- Abscissa: x-coordinate (distance from Y-axis)
- Ordinate: y-coordinate (distance from X-axis)
Cartesian Plane with Four Quadrants
📌 Key Formulae
Distance Formula
Distance between \(P(x_1, y_1)\) and \(Q(x_2, y_2)\):
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Distance from Origin
Distance of \(P(x, y)\) from origin \(= \sqrt{x^2 + y^2}\)
Section Formula (Internal Division)
Point dividing \((x_1, y_1)\) and \((x_2, y_2)\) in ratio \(m:n\):
\[ \left(\frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n}\right) \]
Midpoint Formula
\[ \text{Midpoint} = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \]
📌 Conditions for Geometric Shapes
Using Distance Formula:
- Equilateral Triangle: \(AB = BC = CA\)
- Isosceles Triangle: Any two sides equal
- Right Triangle: \(AB^2 + BC^2 = CA^2\)
- Square: All sides equal AND diagonals equal
- Rhombus: All sides equal, diagonals unequal
- Rectangle: Opposite sides equal, diagonals equal
- Parallelogram: Opposite sides equal, diagonals unequal
Triangles
📌 Similarity of Triangles
Two triangles are similar if:
- Their corresponding angles are equal, AND
- Their corresponding sides are proportional
Notation: \(\triangle ABC \sim \triangle DEF\)
📌 Criteria for Similarity
Four Criteria:
- AAA Criterion: All three corresponding angles are equal
- AA Criterion: Two corresponding angles are equal (third is automatic)
- SSS Criterion: All three corresponding sides are proportional
- SAS Criterion: One angle equal and including sides proportional
📌 Basic Proportionality Theorem (Thales Theorem)
Theorem:
If a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, then the other two sides are divided in the same ratio.
In \(\triangle ABC\), if \(DE \parallel BC\), then:
\[ \frac{AD}{DB} = \frac{AE}{EC} \]
Basic Proportionality Theorem: DE ∥ BC ⟹ AD/DB = AE/EC
📌 Converse of BPT
Theorem:
If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
If \(\dfrac{AD}{DB} = \dfrac{AE}{EC}\), then \(DE \parallel BC\).
🧠 Area of Triangle
"Audi is the product of half of BMW and Honda"
Area = \(\dfrac{1}{2} \times \text{Base} \times \text{Height}\)
Circles
📌 Key Concepts
- Tangent: A line that touches the circle at exactly one point (point of contact)
- Secant: A line intersecting the circle at two distinct points
- The tangent at any point is perpendicular to the radius at the point of contact
- Lengths of tangents drawn from an external point are equal: \(PA = PB\)
- No tangent from a point inside the circle
- Exactly two tangents from a point outside the circle
Tangents from External Point P to Circle with Centre O
Trigonometry and Trigonometric Identities
📌 Trigonometric Ratios
In a right triangle \(ABC\) right-angled at \(B\), with \(\angle BAC = \theta\):
Right Triangle with Trigonometric Sides
Six Trigonometric Ratios
\[ \sin\theta = \frac{P}{H} = \frac{BC}{AC} \]
\[ \cos\theta = \frac{B}{H} = \frac{AB}{AC} \]
\[ \tan\theta = \frac{P}{B} = \frac{BC}{AB} \]
\[ \cot\theta = \frac{B}{P} = \frac{AB}{BC} = \frac{1}{\tan\theta} \]
\[ \sec\theta = \frac{H}{B} = \frac{AC}{AB} = \frac{1}{\cos\theta} \]
\[ \csc\theta = \frac{H}{P} = \frac{AC}{BC} = \frac{1}{\sin\theta} \]
📌 Trigonometric Ratio Table
| Angle | 0° | 30° | 45° | 60° | 90° |
|---|---|---|---|---|---|
| sin | 0 | \(\frac{1}{2}\) | \(\frac{1}{\sqrt{2}}\) | \(\frac{\sqrt{3}}{2}\) | 1 |
| cos | 1 | \(\frac{\sqrt{3}}{2}\) | \(\frac{1}{\sqrt{2}}\) | \(\frac{1}{2}\) | 0 |
| tan | 0 | \(\frac{1}{\sqrt{3}}\) | 1 | \(\sqrt{3}\) | Not defined |
| cot | Not defined | \(\sqrt{3}\) | 1 | \(\frac{1}{\sqrt{3}}\) | 0 |
| sec | 1 | \(\frac{2}{\sqrt{3}}\) | \(\sqrt{2}\) | 2 | Not defined |
| cosec | Not defined | 2 | \(\sqrt{2}\) | \(\frac{2}{\sqrt{3}}\) | 1 |
🧠 Mnemonics for Trigonometric Ratios
"Some People Have" → \(\sin\theta = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}}\)
"Curly Brown Hair" → \(\cos\theta = \dfrac{\text{Base}}{\text{Hypotenuse}}\)
"Through Proper Brushing" → \(\tan\theta = \dfrac{\text{Perpendicular}}{\text{Base}}\)
📌 Trigonometric Identities
Three Fundamental Identities
\[ \sin^2\theta + \cos^2\theta = 1 \]
\[ 1 + \tan^2\theta = \sec^2\theta \]
\[ 1 + \cot^2\theta = \csc^2\theta \]
Heights and Distances
📌 Key Concepts
- Line of Sight: The line from the observer's eye to the point on the object viewed
- Angle of Elevation: Angle formed when looking upward from horizontal
- Angle of Depression: Angle formed when looking downward from horizontal
- Angle of elevation from A to B = Angle of depression from B to A
Angle of Elevation: Observer looking up at an object
📌 Solved Examples
The horizontal distance between two towers is 60 m. The angle of elevation of the top of the taller tower from the top of the shorter one is 30°. If the taller tower is 150 m, find the height of the shorter tower.
Solution: Let height of shorter tower = \(h\) m
Horizontal distance = 60 m, Difference in heights = \(150 - h\)
In the right triangle formed:
\[ \tan 30° = \frac{150 - h}{60} \]
\[ \frac{1}{\sqrt{3}} = \frac{150 - h}{60} \]
\[ 150 - h = \frac{60}{\sqrt{3}} = 20\sqrt{3} \]
\[ h = 150 - 20\sqrt{3} \text{ m} \]
∴ Height of shorter tower = \((150 - 20\sqrt{3})\) m ≈ 115.36 m
If the length of a ladder placed against a wall is twice the distance between the foot of the ladder and the wall, find the angle made by the ladder with the horizontal.
Solution: Let distance from wall = \(x\), then ladder length = \(2x\)
\[ \cos\theta = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{x}{2x} = \frac{1}{2} \]
\[ \cos\theta = \cos 60° \Rightarrow \theta = 60° \]
∴ The angle is 60°.
Areas Related to Circles
📌 Key Formulae
Circle Measurements
Circumference: \(C = 2\pi r = \pi d\)
Area: \(A = \pi r^2\)
Semicircle Area: \(\dfrac{1}{2}\pi r^2\)
Semicircle Perimeter: \(\pi r + 2r = (\pi + 2)r\)
Sector and Segment (Central angle \(\theta\))
Length of Arc: \(l = \dfrac{\theta}{360°} \times 2\pi r\)
Area of Sector: \(A = \dfrac{\theta}{360°} \times \pi r^2 = \dfrac{1}{2}lr\)
Area of Minor Segment: \(\dfrac{\pi r^2 \theta}{360°} - \dfrac{1}{2}r^2\sin\theta\)
Area of Major Segment: \(\pi r^2 - \text{Area of minor segment}\)
Ring (Annulus)
Area of Ring: \(\pi(R^2 - r^2) = \pi(R+r)(R-r)\)
where \(R\) = outer radius, \(r\) = inner radius
Sector (bounded by two radii and arc) vs Segment (bounded by chord and arc)
📌 Special Cases
- If chord subtends 90° at centre: Area of segment = \(r^2\left(\dfrac{\pi}{4} - \dfrac{1}{2}\right)\)
- If chord subtends 60° at centre: Area of segment = \(r^2\left(\dfrac{\pi}{6} - \dfrac{\sqrt{3}}{4}\right)\)
- Distance moved by wheel in 1 revolution = Circumference of wheel
Surface Areas and Volumes
📌 Formulae Summary
| Shape | CSA/LSA | TSA | Volume |
|---|---|---|---|
| Cuboid (l, b, h) |
\(2(l+b)h\) | \(2(lb+bh+hl)\) | \(l \times b \times h\) |
| Cube (edge a) |
\(4a^2\) | \(6a^2\) | \(a^3\) |
| Cylinder (r, h) |
\(2\pi rh\) | \(2\pi r(h+r)\) | \(\pi r^2 h\) |
| Cone (r, h, l) |
\(\pi rl\) | \(\pi r(l+r)\) | \(\dfrac{1}{3}\pi r^2 h\) |
| Sphere (r) |
\(4\pi r^2\) | \(4\pi r^2\) | \(\dfrac{4}{3}\pi r^3\) |
| Hemisphere (r) |
\(2\pi r^2\) | \(3\pi r^2\) | \(\dfrac{2}{3}\pi r^3\) |
Additional Formulae
Slant height of cone: \(l = \sqrt{h^2 + r^2}\)
Diagonal of cuboid: \(\sqrt{l^2 + b^2 + h^2}\)
Diagonal of cube: \(\sqrt{3} \times a\)
Hollow Cylinder TSA: \(2\pi(R+r)(h + R - r)\)
Hollow Cylinder Volume: \(\pi h(R^2 - r^2)\)
Spherical Shell Volume: \(\dfrac{4}{3}\pi(R^3 - r^3)\)
Common 3D Shapes with Key Measurements
🧠 Mnemonics
CSA of Cone: "Pirates really love Sails!" → \(\pi r l\)
CSA of Cylinder: "Two perfect round hats!" → \(2\pi r h\)
Statistics
📌 Measures of Central Tendency
- Mean: Arithmetic average = \(\dfrac{\text{Sum of observations}}{\text{Number of observations}}\)
- Median: Middle value when data is arranged in order
- Mode: Most frequently occurring observation
📌 Mean Formulae
For Raw/Ungrouped Data:
\[ \bar{x} = \frac{x_1 + x_2 + \cdots + x_n}{n} = \frac{1}{n}\sum_{i=1}^{n} x_i \]
\[ \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \]
For Grouped Data — Direct Method:
\[ \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \]
where \(x_i\) = class mark (midpoint), \(f_i\) = frequency
Assumed Mean Method:
\[ \bar{x} = a + \frac{\sum f_i d_i}{\sum f_i} \]
where \(d_i = x_i - a\)
Step-Deviation Method:
\[ \bar{x} = a + h\left(\frac{\sum f_i u_i}{\sum f_i}\right) \]
where \(u_i = \dfrac{x_i - a}{h}\), \(h\) = class size
📌 Median of Grouped Data
\[ \text{Median} = l + \frac{\left(\dfrac{n}{2} - cf\right)}{f} \times h \]
where:
- \(l\) = lower limit of median class
- \(n\) = total number of observations
- \(cf\) = cumulative frequency of class preceding median class
- \(f\) = frequency of median class
- \(h\) = class size
📌 Mode of Grouped Data
\[ \text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h \]
where:
- \(l\) = lower 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
📌 Empirical Relationship
Relationship between Mean, Median, and Mode:
\[ \text{Mode} = 3 \times \text{Median} - 2 \times \text{Mean} \]
\[ \text{Median} = \frac{1}{3}\text{Mode} + \frac{2}{3}\text{Mean} \]
\[ \text{Mean} = \frac{3}{2}\text{Median} - \frac{1}{2}\text{Mode} \]
Probability
📌 Key Concepts
- Random Experiment: Outcome cannot be predicted with certainty (e.g., tossing a coin, throwing a dice)
- Event: Outcome associated with an experiment
- Sure Event: Probability = 1
- Impossible Event: Probability = 0
- Elementary Event: Event with only one outcome
- Probability always lies between 0 and 1: \(0 \leq P(E) \leq 1\)
📌 Probability Formula
Theoretical Probability
\[ P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \]
Complementary Event
\[ P(E) + P(\bar{E}) = 1 \quad \text{or} \quad P(\bar{E}) = 1 - P(E) \]
Sum of all elementary events = 1
📌 Playing Cards
A Standard Deck has 52 Cards:
- 4 Suits: ♠ Spades, ♥ Hearts, ♦ Diamonds, ♣ Clubs
- Red suits: Hearts ♥ and Diamonds ♦ (26 cards)
- Black suits: Spades ♠ and Clubs ♣ (26 cards)
- Each suit has: Ace, King, Queen, Jack, 2-10 (13 cards)
- Face cards: King, Queen, Jack (12 total, 3 per suit)
Standard Deck of 52 Playing Cards
📌 Important Facts
- The first book on probability, "The Book on Games of Chance," was written by Italian mathematician J. Cardan
- The classical definition of probability was given by Pierre Simon Laplace
- Experimental probability is based on actual trials; theoretical probability is based on assumptions
- As number of trials increases, experimental probability approaches theoretical probability
- A "fair" coin is symmetrical and unbiased
📌 Common Probability Questions
Quick Reference:
- P(Head on coin toss) = \(\dfrac{1}{2}\)
- P(getting 6 on dice) = \(\dfrac{1}{6}\)
- P(drawing Ace from deck) = \(\dfrac{4}{52} = \dfrac{1}{13}\)
- P(drawing red card) = \(\dfrac{26}{52} = \dfrac{1}{2}\)
- P(drawing face card) = \(\dfrac{12}{52} = \dfrac{3}{13}\)