Class 10 (SSC Board) · Mathematics Part 1 (Algebra) · Chapter 1
Linear Equations in Two Variables
Quick summary
Two equations in x and y are solved together to find one pair (x, y) that fits both. You can solve them by graph, substitution, elimination or Cramer's rule.
Important terms
Linear equation in two variables – ax + by = c, where a and b are not both 0.
Simultaneous equations – two equations solved together.
Determinant – a number made from 4 numbers: |a b; c d| = ad − bc.
Formulas
For a₁x + b₁y = c₁ and a₂x + b₂y = c₂ (Cramer's rule):
D = a₁b₂ − a₂b₁
Dx = c₁b₂ − c₂b₁
Dy = a₁c₂ − a₂c₁
x = Dx/D, y = Dy/D (only when D ≠ 0)
Solved examples
Example 1. Solve: x + y = 5 and x − y = 1 (elimination)
Add both: 2x = 6, so x = 3. Put x = 3 in x + y = 5: y = 2. Answer: x = 3, y = 2
Example 2. Solve by Cramer's rule: 3x − 4y = 10 and 4x + 3y = 5
D = (3)(3) − (4)(−4) = 9 + 16 = 25 Dx = (10)(3) − (5)(−4) = 30 + 20 = 50 Dy = (3)(5) − (4)(10) = 15 − 40 = −25 x = 50/25 = 2, y = −25/25 = −1 Answer: x = 2, y = −1
Quick revision
Always write both equations in the form ax + by = c first.
Elimination: make the x (or y) coefficients equal, then add or subtract.
Graph method: the meeting point of the two lines is the solution.
If D = 0, Cramer's rule cannot be used.
Equations like 1/x and 1/y: put 1/x = m, 1/y = n, solve, then go back to x and y.
Word problems: choose the two unknowns, make two equations, solve.
Exam tip
Check your answer by putting x and y back into both equations. It takes 10 seconds and saves marks.
Class 10 (SSC Board) · Mathematics Part 1 (Algebra) · Chapter 2
Quadratic Equations
Quick summary
A quadratic equation has the highest power 2. It has two roots. You find them by factorisation, completing the square or the formula, and the discriminant tells you the type of roots.
Important terms
Quadratic equation – ax² + bx + c = 0, where a ≠ 0.
Root – a value of x that makes the equation true.
Discriminant (Δ) – b² − 4ac.
Formulas
Roots: x = [−b ± √(b² − 4ac)] / 2a
Δ = b² − 4ac
Sum of roots: α + β = −b/a
Product of roots: αβ = c/a
Equation from roots: x² − (α + β)x + αβ = 0
Solved examples
Example 1. Solve x² − 5x + 6 = 0 by factorisation
Split −5x as −2x − 3x: x² − 2x − 3x + 6 = 0 x(x − 2) − 3(x − 2) = 0 → (x − 2)(x − 3) = 0 Roots: x = 2 or x = 3
Example 2. Solve 2x² + 5x − 3 = 0 by formula
a = 2, b = 5, c = −3 Δ = 25 − 4(2)(−3) = 25 + 24 = 49, √Δ = 7 x = (−5 ± 7)/4 → x = 2/4 or x = −12/4 Roots: x = 1/2 or x = −3
Quick revision
Δ > 0 → roots real and unequal.
Δ = 0 → roots real and equal.
Δ < 0 → roots not real.
To check if a value is a root, put it in the equation; the answer must be 0.
Completing the square: add (half of the x-coefficient)² on both sides (after making a = 1).
Always write the equation in the form ax² + bx + c = 0 before finding a, b, c.
Exam tip
Write a, b, c with their signs first. Most mistakes happen with the minus sign of b or c.
Class 10 (SSC Board) · Mathematics Part 1 (Algebra) · Chapter 3
Arithmetic Progression
Quick summary
An A.P. is a list of numbers where you add the same number every time to get the next one. This chapter teaches how to find any term (tₙ) and the sum of terms (Sₙ).
Important terms
Sequence – numbers written in a fixed order: t₁, t₂, t₃, ...
Arithmetic Progression (A.P.) – a sequence where the difference between any two consecutive terms is the same.
Common difference (d) – d = t₂ − t₁ = t₃ − t₂ = ...
Formulas
nth term: tₙ = a + (n − 1)d
Sum of first n terms: Sₙ = n/2 [2a + (n − 1)d]
Sₙ = n/2 (first term + last term)
tₙ = Sₙ − Sₙ₋₁
(a = first term, d = common difference, n = number of terms)
Solved examples
Example 1. Find the 15th term of the A.P. 3, 8, 13, ...
a = 3, d = 8 − 3 = 5, n = 15 t₁₅ = 3 + (15 − 1) × 5 = 3 + 70 t₁₅ = 73
Example 2. Find the sum of the first 20 terms of 2, 6, 10, ...
Class 10 (SSC Board) · Mathematics Part 1 (Algebra) · Chapter 6
Statistics
Quick summary
Statistics is about grouped data. You find mean, median and mode from a frequency table and show data with graphs.
Important terms
Class mark (xᵢ) – middle of a class = (lower limit + upper limit)/2.
Frequency (fᵢ) – how many times values fall in a class.
Cumulative frequency (cf) – running total of frequencies.
Median class – the class where N/2 lies.
Modal class – the class with the highest frequency.
Formulas
Mean (direct): x̄ = Σfᵢxᵢ / Σfᵢ
Mean (assumed mean): x̄ = A + Σfᵢdᵢ / Σfᵢ, where dᵢ = xᵢ − A
Mean (step deviation): x̄ = A + (Σfᵢuᵢ / Σfᵢ) × g, where uᵢ = (xᵢ − A)/g
Median = L + [(N/2 − cf) / f] × h
Mode = L + [(f₁ − f₀) / (2f₁ − f₀ − f₂)] × h
Pie diagram: central angle = (value / total) × 360°
(L = lower limit of the class, h = class width, f = frequency of the class, cf = cumulative frequency before it, f₁ = modal class frequency, f₀ = class before, f₂ = class after)
Solved examples
Use this table for both examples:
Class
0–10
10–20
20–30
30–40
40–50
Frequency
5
8
12
10
5
Example 1. Find the median.
N = 40, N/2 = 20. cf: 5, 13, 25, ... so the median class is 20–30. L = 20, cf = 13, f = 12, h = 10 Median = 20 + [(20 − 13)/12] × 10 = 20 + 5.83 Median ≈ 25.83