Chapters
🧮 Mathematics Basic — Class X
Simpler questions, same chapters. Focus on concepts and application.
80 Marks
3 Hours
Internal: 20M
Easier than Standard
📌 Maths Basic vs Standard: Same chapters, same NCERT textbook — but Basic paper has simpler questions, more direct application, less proof-based questions. Ideal for non-Maths stream in XI-XII.
15
Chapters
80
Board Marks
20
Internal Marks
Number Systems
Real Numbers, HCF, LCM
Algebra
Polynomials, Equations, AP
Geometry
Triangles, Circles, Areas
Trigonometry
Ratios, Heights & Distances
Statistics
Mean, Median, Mode, Prob.
🏅 Badges
🔢 Real Numbers
Ch 1 · Euclid's Division · FTA · Irrational Numbers · Decimal Expansions
📖 Key Concepts
Euclid's Division Lemma
For any two positive integers a and b, unique integers q and r exist such that a = bq + r, 0 ≤ r < b. Used to find HCF step-by-step.
HCF by Euclid
Divide larger by smaller; replace with (smaller, remainder); repeat until remainder = 0. Last non-zero remainder = HCF(a, b).
Fundamental Theorem of Arithmetic
Every composite number is a unique product of primes (order aside). Forms basis for LCM and HCF via prime factorisation.
LCM × HCF = a × b
Valid for exactly two numbers. If HCF = 12 and LCM = 360, product = 12 × 360 = 4320 = a × b.
Irrational Numbers
Cannot be written as p/q (q ≠ 0). Decimal is non-terminating, non-repeating. √2, √3, √5, π, e are all irrational.
Terminating Decimal
p/q terminates ⟺ q = 2m × 5n. If denominator has any other prime factor, decimal is non-terminating repeating.
🔑 Master Formulae
Euclid: a = bq + r, 0 ≤ r < b
HCF(a,b) × LCM(a,b) = a × b (two numbers only)
Terminating: denominator = 2m · 5n only
√p is irrational for any prime p (proof by contradiction)
✏️ Worked — HCF by Euclid (Board 2023, 2M)
Q: Find HCF(252, 105).
1
252 = 105 × 2 + 42
2
105 = 42 × 2 + 21
3
42 = 21 × 2 + 0 → stop
✅ HCF(252, 105) = 21
✏️ Worked — Decimal Type (Board 2022, 1M)
Q: Is 17/8 terminating or non-terminating?
1
Denominator 8 = 2³. No prime other than 2.
✅ Terminating. 17/8 = 2.125
📝 Chapter MCQs
MCQ 1 HCF(8, 9, 25) = ?
Options: (A) 1 (B) 2 (C) 3 (D) 5
Ans: (A) 1 — all three are co-prime with each other.
Options: (A) 1 (B) 2 (C) 3 (D) 5
Ans: (A) 1 — all three are co-prime with each other.
MCQ 2 Which of these is irrational?
(A) √9 (B) √25 (C) √7 (D) √16
Ans: (C) √7 — 7 is prime, so √7 is irrational.
(A) √9 (B) √25 (C) √7 (D) √16
Ans: (C) √7 — 7 is prime, so √7 is irrational.
MCQ 3 The decimal expansion of 7/20 is:
(A) Terminating (B) Non-terminating recurring (C) Non-recurring (D) None
Ans: (A) — 20 = 2² × 5, so it terminates. 7/20 = 0.35.
(A) Terminating (B) Non-terminating recurring (C) Non-recurring (D) None
Ans: (A) — 20 = 2² × 5, so it terminates. 7/20 = 0.35.
MCQ 4 If HCF(a, b) = 12 and a × b = 1800, LCM = ?
(A) 100 (B) 150 (C) 216 (D) 300
Ans: (B) 150 — LCM = 1800 ÷ 12 = 150.
(A) 100 (B) 150 (C) 216 (D) 300
Ans: (B) 150 — LCM = 1800 ÷ 12 = 150.
MCQ 5 √2 × √8 is:
(A) Rational (B) Irrational (C) Integer (D) Fraction
Ans: (A) Rational — √2 × √8 = √16 = 4, a rational number.
(A) Rational (B) Irrational (C) Integer (D) Fraction
Ans: (A) Rational — √2 × √8 = √16 = 4, a rational number.
🔁 Assertion–Reason
A-R
Assertion (A): √5 is irrational.
Reason (R): Square root of any positive integer is irrational.
Options: (A) Both A and R are true, R is the correct explanation (B) Both true, R is NOT correct explanation (C) A true, R false (D) A false, R true
Ans: (C) — A is true (√5 is irrational). R is false — √4 = 2 is rational.
Reason (R): Square root of any positive integer is irrational.
Options: (A) Both A and R are true, R is the correct explanation (B) Both true, R is NOT correct explanation (C) A true, R false (D) A false, R true
Ans: (C) — A is true (√5 is irrational). R is false — √4 = 2 is rational.
A-R
Assertion (A): 6/15 has terminating decimal expansion.
Reason (R): A fraction p/q terminates if q = 2m × 5n.
Ans: (A) — 6/15 = 2/5; denominator 5 = 5¹ ✓. R correctly explains A.
Reason (R): A fraction p/q terminates if q = 2m × 5n.
Ans: (A) — 6/15 = 2/5; denominator 5 = 5¹ ✓. R correctly explains A.
📦 Case-Based Question
Passage: A school timetable committee is scheduling periods for Music (every 36 minutes) and Sports (every 54 minutes). Both start at 8:00 AM. The committee also wants to know if the total minutes in the school day (360 min) can be written as a product of primes.
Q1 (1M) Find HCF(36, 54). Ans: 18 (36 = 2²×3², 54 = 2×3³; HCF = 2×3² = 18)
Q2 (1M) When will both activities next coincide? Ans: LCM = 36×54/18 = 108 min = 9:48 AM.
Q3 (2M) Write prime factorisation of 360. Ans: 360 = 2³ × 3² × 5.
✍️ Short Answer (2–3 Marks)
SA 1 (2M) Prove that √3 is irrational.
Ans: Assume √3 = p/q (lowest terms). Then 3 = p²/q², so p² = 3q². Thus 3 | p², so 3 | p. Let p = 3m. Then 9m² = 3q², so q² = 3m², meaning 3 | q. But then 3 is a common factor of p and q, contradicting our assumption. ∴ √3 is irrational. ✓
Ans: Assume √3 = p/q (lowest terms). Then 3 = p²/q², so p² = 3q². Thus 3 | p², so 3 | p. Let p = 3m. Then 9m² = 3q², so q² = 3m², meaning 3 | q. But then 3 is a common factor of p and q, contradicting our assumption. ∴ √3 is irrational. ✓
SA 2 (2M) Find HCF and LCM of 510 and 92 using prime factorisation.
Ans: 510 = 2 × 3 × 5 × 17; 92 = 2² × 23. HCF = 2. LCM = 2² × 3 × 5 × 17 × 23 = 23460.
Ans: 510 = 2 × 3 × 5 × 17; 92 = 2² × 23. HCF = 2. LCM = 2² × 3 × 5 × 17 × 23 = 23460.
📘 Long Answer (5 Marks)
LA (5M) Three bells ring at intervals of 6, 12, and 18 minutes. All ring together at 8:00 AM. (i) After how many minutes will they ring together again? (ii) How many times will they ring together in next 3 hours? (iii) Is this interval rational or irrational — justify.
Ans: (i) LCM(6,12,18) = 36 minutes. (ii) 3 hours = 180 min; 180 ÷ 36 = 5 times (not counting 8:00 AM start). (iii) 36 is rational — it equals 36/1, which is p/q form with q ≠ 0.
Ans: (i) LCM(6,12,18) = 36 minutes. (ii) 3 hours = 180 min; 180 ÷ 36 = 5 times (not counting 8:00 AM start). (iii) 36 is rational — it equals 36/1, which is p/q form with q ≠ 0.
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🧮 Polynomials
Ch 2 · Zeroes · Relationship with Coefficients · Division Algorithm
📖 Key Concepts
Zero of a Polynomial
Value of x for which p(x) = 0. Linear p(x) has 1 zero; quadratic has at most 2; cubic has at most 3.
Sum & Product of Zeroes
For ax² + bx + c: α + β = −b/a; α × β = c/a. Key shortcuts — memorise both!
Geometrical Meaning
Zero = x-intercept of the graph. Number of zeroes = number of times graph cuts x-axis.
Division Algorithm
p(x) = g(x) × q(x) + r(x), where degree(r) < degree(g). Same as Euclid but for polynomials.
🔑 Quadratic Zeroes Formulae
α + β = −b/a | αβ = c/a
Quadratic with zeroes α, β: x² − (α+β)x + αβ = 0
If zeroes are given, form: k[x² − (sum)x + (product)]
✏️ Worked — Find zeroes (Board 2023, 2M)
Q: Find zeroes of p(x) = x² − 5x + 6 and verify.
1
Factor: x² − 5x + 6 = (x−2)(x−3). Zeroes: x = 2 and x = 3.
2
Verify: α+β = 2+3 = 5 = −(−5)/1 ✓ ; αβ = 2×3 = 6 = 6/1 ✓
✅ Zeroes are 2 and 3.
📝 Chapter MCQs
MCQ 1 Zeroes of x² − 3x − 4 are:
(A) −1, 4 (B) 1, −4 (C) −1, −4 (D) 1, 4
Ans: (A) — (x−4)(x+1)=0.
(A) −1, 4 (B) 1, −4 (C) −1, −4 (D) 1, 4
Ans: (A) — (x−4)(x+1)=0.
MCQ 2 If sum of zeroes = 3 and product = −10, the quadratic is:
(A) x²+3x−10 (B) x²−3x+10 (C) x²−3x−10 (D) x²+3x+10
Ans: (C) x²−(sum)x+(product) = x²−3x−10.
(A) x²+3x−10 (B) x²−3x+10 (C) x²−3x−10 (D) x²+3x+10
Ans: (C) x²−(sum)x+(product) = x²−3x−10.
MCQ 3 Number of zeroes of a cubic polynomial (at most):
(A) 1 (B) 2 (C) 3 (D) 4
Ans: (C) 3
(A) 1 (B) 2 (C) 3 (D) 4
Ans: (C) 3
🔁 Assertion–Reason
A-R
A: x² + 1 has no real zeroes.
R: The discriminant b²−4ac = 0−4 = −4 < 0.
Ans: (A) Both true, R correctly explains A.
R: The discriminant b²−4ac = 0−4 = −4 < 0.
Ans: (A) Both true, R correctly explains A.
📦 Case-Based Question
Passage: A garden path is rectangular. Its area is represented by p(x) = x² + 5x + 6 square metres, where x is a positive integer.
Q1 Find zeroes of p(x). Ans: (x+2)(x+3)=0; zeroes −2 and −3.
Q2 What are the possible dimensions if x = 4? Ans: p(4) = 16+20+6 = 42 m². Dimensions: (4+2)(4+3) = 6m × 7m.
Q3 Verify sum and product of zeroes. Ans: Sum = −2+(−3) = −5 = −5/1 ✓; Product = (−2)(−3) = 6 = 6/1 ✓
✍️ Short Answer
SA (3M) If α and β are zeroes of 2x² + 5x − 12, find α + β, αβ, and 1/α + 1/β.
Ans: α+β = −5/2; αβ = −12/2 = −6. 1/α + 1/β = (α+β)/(αβ) = (−5/2)/(−6) = 5/12.
Ans: α+β = −5/2; αβ = −12/2 = −6. 1/α + 1/β = (α+β)/(αβ) = (−5/2)/(−6) = 5/12.
⚖️ Pair of Linear Equations
Ch 3 · Graphical · Substitution · Elimination · Cross-Multiplication
📖 Key Concepts
Consistent System
a₁/a₂ ≠ b₁/b₂ → unique solution (lines intersect). a₁/a₂ = b₁/b₂ = c₁/c₂ → infinitely many solutions (coincident lines).
Inconsistent System
a₁/a₂ = b₁/b₂ ≠ c₁/c₂ → no solution (lines are parallel). Graph shows parallel lines that never meet.
Substitution Method
Express one variable from one equation; substitute into the other. Best when a coefficient is 1.
Elimination Method
Multiply equations to make coefficients equal; add/subtract to eliminate one variable. Most reliable method.
🔑 Consistency Conditions
Unique solution: a₁/a₂ ≠ b₁/b₂
No solution (parallel): a₁/a₂ = b₁/b₂ ≠ c₁/c₂
Infinite solutions (same line): a₁/a₂ = b₁/b₂ = c₁/c₂
✏️ Worked — Elimination (Board 2023, 3M)
Q: Solve: 2x + 3y = 11 and 2x − 4y = −24.
1
Subtract eq2 from eq1: 7y = 35 → y = 5.
2
Substitute: 2x + 15 = 11 → 2x = −4 → x = −2.
✅ x = −2, y = 5
📝 Chapter MCQs
MCQ 1 x + y = 5 and 2x + 2y = 10 represents:
(A) unique solution (B) no solution (C) infinitely many (D) none
Ans: (C) — ratios 1/2 = 1/2 = 5/10, same line.
(A) unique solution (B) no solution (C) infinitely many (D) none
Ans: (C) — ratios 1/2 = 1/2 = 5/10, same line.
MCQ 2 Graphically, 3x − y = 2 and 9x − 3y = 12 are:
(A) intersecting (B) coincident (C) parallel (D) perpendicular
Ans: (C) Parallel — 3/9 = 1/3 = −1/−3, but 2/12 ≠ 1/3.
(A) intersecting (B) coincident (C) parallel (D) perpendicular
Ans: (C) Parallel — 3/9 = 1/3 = −1/−3, but 2/12 ≠ 1/3.
🔁 Assertion–Reason
A-R
A: x + y = 6 and 2x + 2y = 12 have infinitely many solutions.
R: The two equations represent the same line.
Ans: (A) Both correct, R explains A.
R: The two equations represent the same line.
Ans: (A) Both correct, R explains A.
📦 Case-Based
Passage: Two friends Riya and Priya together have 45 marbles. Riya has 15 more than Priya. Let Riya = x, Priya = y.
Q1 Form the equations. Ans: x + y = 45; x − y = 15.
Q2 Solve for x and y. Ans: Adding: 2x = 60; x = 30, y = 15.
Q3 What type of system is this? Ans: Consistent with unique solution (lines intersect).
✍️ Short & Long Answer
SA (3M) Solve by substitution: 3x + 2y = 10 and x − y = 1.
Ans: x = 1 + y → 3(1+y) + 2y = 10 → 5y = 7 → y = 7/5; x = 1 + 7/5 = 12/5.
Ans: x = 1 + y → 3(1+y) + 2y = 10 → 5y = 7 → y = 7/5; x = 1 + 7/5 = 12/5.
LA (5M) A boat covers 32 km upstream and 36 km downstream in 7 hours. It covers 40 km upstream and 48 km downstream in 9 hours. Find speed of boat and stream.
Ans: Let boat speed = b, stream = s. Up-speed = b−s, down-speed = b+s. Equations: 32/(b−s) + 36/(b+s) = 7 and 40/(b−s) + 48/(b+s) = 9. Let u = 1/(b−s), v = 1/(b+s). Solve: u = 1/8, v = 1/12. So b−s = 8, b+s = 12. Therefore b = 10 km/h, s = 2 km/h.
Ans: Let boat speed = b, stream = s. Up-speed = b−s, down-speed = b+s. Equations: 32/(b−s) + 36/(b+s) = 7 and 40/(b−s) + 48/(b+s) = 9. Let u = 1/(b−s), v = 1/(b+s). Solve: u = 1/8, v = 1/12. So b−s = 8, b+s = 12. Therefore b = 10 km/h, s = 2 km/h.
📐 Quadratic Equations
Ch 4 · Factorisation · Quadratic Formula · Discriminant · Nature of Roots
📖 Key Concepts
Standard Form
ax² + bx + c = 0, a ≠ 0. Every quadratic has at most two real roots.
Quadratic Formula
x = [−b ± √(b²−4ac)] / 2a. Always works even when factorisation is hard.
Discriminant (D)
D = b²−4ac. D > 0: two distinct real roots. D = 0: two equal roots. D < 0: no real roots.
Completing the Square
Write ax²+bx+c = a(x+b/2a)² + (c−b²/4a). Useful method to derive the formula.
🔑 Key Formulae
x = [−b ± √(b²−4ac)] / 2a
D = b²−4ac: D>0 distinct; D=0 equal; D<0 no real roots
For equal roots: D = 0 → b² = 4ac
✏️ Worked — Discriminant (Board 2023, 3M)
Q: Find the nature of roots: 2x² − 4x + 3 = 0.
1
D = (−4)² − 4(2)(3) = 16 − 24 = −8.
2
D < 0 → no real roots.
✅ No real roots exist.
📝 MCQs + A-R
MCQ 1 Which value of k gives equal roots for kx² + 2x + 1 = 0?
(A) k=1 (B) k=2 (C) k=−1 (D) k=0
Ans: (A) k=1 — D=4−4k=0 → k=1.
(A) k=1 (B) k=2 (C) k=−1 (D) k=0
Ans: (A) k=1 — D=4−4k=0 → k=1.
MCQ 2 Roots of x² − 7x + 10 = 0 are:
(A) 2, 5 (B) −2, −5 (C) 2, −5 (D) −2, 5
Ans: (A) 2, 5
(A) 2, 5 (B) −2, −5 (C) 2, −5 (D) −2, 5
Ans: (A) 2, 5
A-R
A: x² + 1 = 0 has no solution in real numbers.
R: D = 0 − 4(1)(1) = −4 < 0.
Ans: (A) Both correct, R correctly explains.
R: D = 0 − 4(1)(1) = −4 < 0.
Ans: (A) Both correct, R correctly explains.
📦 Case-Based + SA + LA
Passage: A rectangular field has length 4 more than breadth. Area = 96 m². Let breadth = x.
Q1 Form the equation. Ans: x(x+4) = 96 → x² + 4x − 96 = 0.
Q2 Find the dimensions. Ans: (x+12)(x−8) = 0 → x = 8 (positive). Breadth=8m, Length=12m.
SA (2M) Find k if 3x² − kx + 4 = 0 has equal roots.
Ans: D = k² − 48 = 0 → k = ±4√3.
Ans: D = k² − 48 = 0 → k = ±4√3.
LA (5M) A train travels 360 km at uniform speed. If speed were 5 km/h more, time would be 1 hr less. Find speed.
Ans: Speed = x; 360/x − 360/(x+5) = 1 → 360(x+5) − 360x = x(x+5) → 1800 = x²+5x → x²+5x−1800=0 → (x+45)(x−40)=0 → x = 40 km/h.
Ans: Speed = x; 360/x − 360/(x+5) = 1 → 360(x+5) − 360x = x(x+5) → 1800 = x²+5x → x²+5x−1800=0 → (x+45)(x−40)=0 → x = 40 km/h.
🔢 Arithmetic Progressions
Ch 5 · nth Term · Sum of n Terms · Identifying APs
📖 Key Concepts
Arithmetic Progression
Sequence where difference between consecutive terms is constant. Common difference d = a₂ − a₁ = a₃ − a₂.
nth Term (aₙ)
aₙ = a + (n−1)d, where a = first term, d = common difference. Last term l = a + (n−1)d.
Sum of n Terms (Sₙ)
Sₙ = n/2 × [2a + (n−1)d] = n/2 × (first + last term). Also: nth term = Sₙ − Sₙ₋₁.
Three Terms in AP
Take as (a−d), a, (a+d). Four terms: (a−3d), (a−d), (a+d), (a+3d). Simplifies calculations.
🔑 AP Formulae
aₙ = a + (n−1)d
Sₙ = n/2 [2a + (n−1)d] = n/2 (a + l)
d = (l−a)/(n−1)
nth term from end = l − (n−1)d
✏️ Worked — Sum problem (Board 2022, 4M)
Q: Find number of terms in AP 3, 7, 11... whose sum = 406.
1
a=3, d=4. Sₙ = n/2[6+(n−1)4] = n(2n+1).
2
n(2n+1)=406 → 2n²+n−406=0 → n=14 (taking positive root).
✅ 14 terms needed.
📝 MCQs + A-R
MCQ 1 10th term of AP: −5, −1, 3, 7... is:
(A) 31 (B) 35 (C) 27 (D) 39
Ans: (A) 31 — a=−5, d=4; a₁₀ = −5+36 = 31.
(A) 31 (B) 35 (C) 27 (D) 39
Ans: (A) 31 — a=−5, d=4; a₁₀ = −5+36 = 31.
MCQ 2 Sum of first 10 natural numbers is:
(A) 45 (B) 55 (C) 50 (D) 60
Ans: (B) 55 — S₁₀ = 10/2 × 11 = 55.
(A) 45 (B) 55 (C) 50 (D) 60
Ans: (B) 55 — S₁₀ = 10/2 × 11 = 55.
A-R
A: 2, 4, 8, 16 is NOT an AP.
R: Common difference between consecutive terms is not constant (2, 4, 8...).
Ans: (A) Both correct, R explains A.
R: Common difference between consecutive terms is not constant (2, 4, 8...).
Ans: (A) Both correct, R explains A.
📦 Case + SA + LA
Passage: A student saves ₹1 on day 1, ₹2 on day 2, ₹3 on day 3, continuing for 30 days (an AP).
Q1 How much on day 30? Ans: a₃₀ = 1 + 29×1 = ₹30.
Q2 Total savings? Ans: S₃₀ = 30/2 × (1+30) = ₹465.
Q3 On which day does saving cross ₹25? Ans: aₙ = n > 25 → day 26.
SA (3M) In an AP, 7th term is 13 and 11th term is 21. Find the AP.
Ans: a₇ = a+6d = 13; a₁₁ = a+10d = 21. Subtract: 4d = 8 → d = 2. a = 13−12 = 1. AP: 1, 3, 5, 7...
Ans: a₇ = a+6d = 13; a₁₁ = a+10d = 21. Subtract: 4d = 8 → d = 2. a = 13−12 = 1. AP: 1, 3, 5, 7...
🔺 Triangles
Ch 6 · Similarity · Pythagoras Theorem · Criteria AA/SAS/SSS
📖 Key Concepts
Basic Proportionality Theorem
If a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally. (Thales' Theorem)
Similarity Criteria
AA: two angles equal. SAS: two sides proportional with included angle equal. SSS: all three sides proportional.
Areas of Similar Triangles
Ratio of areas = square of ratio of corresponding sides. (area₁/area₂ = (side₁/side₂)²)
Pythagoras Theorem
In right △, (hypotenuse)² = sum of squares of other two sides. Converse: if a²+b²=c², it is a right triangle.
🔑 Key Results
BPT: DE ∥ BC ⟹ AD/DB = AE/EC
Area ratio: Δ₁/Δ₂ = (AB/PQ)² = (BC/QR)² = (CA/RP)²
Pythagoras: AC² = AB² + BC² (right angle at B)
✏️ Worked — BPT (Board 2023, 3M)
Q: In △ABC, DE ∥ BC. AD = 3, DB = 5, AE = x, EC = 10. Find x.
1
By BPT: AD/DB = AE/EC → 3/5 = x/10.
2
x = 30/5 = 6.
✅ AE = 6 units
📝 MCQs + A-R
MCQ 1 If △ABC ~ △PQR and BC/QR = 3/5, ratio of areas is:
(A) 3/5 (B) 6/10 (C) 9/25 (D) 27/125
Ans: (C) 9/25
(A) 3/5 (B) 6/10 (C) 9/25 (D) 27/125
Ans: (C) 9/25
MCQ 2 A 12m ladder leans against a wall reaching 8m. Distance from base to wall:
(A) 8.94m (B) √80m (C) 4√5m (D) Both B and C
Ans: (D) — √(144−64) = √80 = 4√5.
(A) 8.94m (B) √80m (C) 4√5m (D) Both B and C
Ans: (D) — √(144−64) = √80 = 4√5.
A-R
A: All squares are similar.
R: All corresponding angles are equal (90°) and corresponding sides are proportional.
Ans: (A)
R: All corresponding angles are equal (90°) and corresponding sides are proportional.
Ans: (A)
📦 Case + SA + LA
Passage: Two poles of height 6m and 11m stand on level ground. Wires connect their tops to the base of the other pole. Point of intersection of wires is at height h from ground.
Q1 This involves similar triangles concept. If poles are 10m apart, draw and identify the triangles. Ans: Two pairs of similar triangles formed by the crossing wires.
Q2 Height of intersection = (6×11)/(6+11) ≈ 3.9 m. Verify this is less than both poles. Ans: Yes, 3.9 < 6 < 11. ✓
LA (5M) Prove Pythagoras theorem. State its converse and use it to check if a triangle with sides 8, 15, 17 is right-angled.
Ans (outline): Standard proof using similar triangles (construct altitude to hypotenuse). Converse: if c²=a²+b², it's right-angled. Check: 8²+15²=64+225=289=17². ✓ Right-angled at the angle between 8 and 15.
Ans (outline): Standard proof using similar triangles (construct altitude to hypotenuse). Converse: if c²=a²+b², it's right-angled. Check: 8²+15²=64+225=289=17². ✓ Right-angled at the angle between 8 and 15.
🗺️ Coordinate Geometry
Ch 7 · Distance Formula · Section Formula · Area of Triangle
📖 Key Concepts
Distance Formula
Distance PQ = √[(x₂−x₁)² + (y₂−y₁)²]. Always positive. Used to show collinearity, equidistance, etc.
Section Formula
Point dividing P(x₁,y₁) and Q(x₂,y₂) in ratio m:n internally: x = (mx₂+nx₁)/(m+n), y = (my₂+ny₁)/(m+n).
Midpoint Formula
Special case of section (1:1): midpoint = ((x₁+x₂)/2, (y₁+y₂)/2).
Area of Triangle
Area = ½ |x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)|. Zero means points are collinear.
🔑 Key Formulae
Distance = √[(x₂−x₁)² + (y₂−y₁)²]
Section (m:n): x = (mx₂+nx₁)/(m+n)
Midpoint = ((x₁+x₂)/2, (y₁+y₂)/2)
Area = ½|x₁(y₂−y₃)+x₂(y₃−y₁)+x₃(y₁−y₂)|
✏️ Worked — Distance (Board 2022, 2M)
Q: Show that A(3,−2), B(1,0), C(−1,2) are collinear.
1
AB = √[(3−1)²+(−2−0)²] = √8 = 2√2
2
BC = √[(1+1)²+(0−2)²] = √8 = 2√2. AC = √[(3+1)²+(−2−2)²] = √32 = 4√2.
3
AB + BC = 2√2 + 2√2 = 4√2 = AC ✓ → collinear.
✅ Points are collinear.
📝 MCQs + A-R + Case
MCQ 1 Midpoint of (3,4) and (−1,2) is:
(A) (1,3) (B) (2,6) (C) (1,2) (D) (−1,3)
Ans: (A) (1,3)
(A) (1,3) (B) (2,6) (C) (1,2) (D) (−1,3)
Ans: (A) (1,3)
MCQ 2 Distance of (3,4) from origin:
(A) 25 (B) 5 (C) 7 (D) 12
Ans: (B) 5 — √(9+16)=5.
(A) 25 (B) 5 (C) 7 (D) 12
Ans: (B) 5 — √(9+16)=5.
A-R
A: (0,0), (1,1), (2,2) are collinear.
R: Area of triangle formed = 0.
Ans: (A) Both true, R correctly explains A.
R: Area of triangle formed = 0.
Ans: (A) Both true, R correctly explains A.
Passage: A city map has three bus stops at A(1,2), B(7,5), C(4,−1). A new stop D divides AB in ratio 2:1.
Q1 Find D. Ans: D = ((2×7+1×1)/3, (2×5+1×2)/3) = (5, 4).
Q2 Find AC. Ans: AC = √[(1−4)²+(2+1)²] = √(9+9) = 3√2.
SA (3M) Find area of △ with vertices (2,3), (−1,0), (2,−4).
Ans: Area = ½|2(0+4)+(−1)(−4−3)+2(3−0)| = ½|8+7+6| = ½×21 = 10.5 sq units.
Ans: Area = ½|2(0+4)+(−1)(−4−3)+2(3−0)| = ½|8+7+6| = ½×21 = 10.5 sq units.
📏 Introduction to Trigonometry
Ch 8 · Trig Ratios · Standard Angles · Complementary Angles · Identities
📖 Key Concepts
Six Trig Ratios
sin A = P/H; cos A = B/H; tan A = P/B; cosec A = H/P; sec A = H/B; cot A = B/P. (P=perpendicular, B=base, H=hypotenuse)
Standard Angles
0°, 30°, 45°, 60°, 90°. sin grows 0→1; cos decreases 1→0 as angle increases from 0° to 90°.
Complementary Angles
sin(90°−A) = cos A; cos(90°−A) = sin A; tan(90°−A) = cot A. Used to simplify expressions.
Pythagorean Identities
sin²A + cos²A = 1 (most important!); 1 + tan²A = sec²A; 1 + cot²A = cosec²A.
🔑 Standard Values Table
sin: 0°=0, 30°=½, 45°=1/√2, 60°=√3/2, 90°=1
cos: 0°=1, 30°=√3/2, 45°=1/√2, 60°=½, 90°=0
tan: 0°=0, 30°=1/√3, 45°=1, 60°=√3, 90°=∞
Identities: sin²+cos²=1; sec²−tan²=1; cosec²−cot²=1
✏️ Worked — Identity Proof (Board 2023, 3M)
Q: Prove: (sin A − 2sin³A)/(2cos³A − cos A) = tan A.
1
LHS = sinA(1−2sin²A) / cosA(2cos²A−1)
2
= sinA(1−2sin²A) / cosA(2(1−sin²A)−1) = sinA(1−2sin²A) / cosA(1−2sin²A)
3
= sinA / cosA = tanA = RHS ✓
✅ Proved.
📝 MCQs + A-R
MCQ 1 sin 60° × cos 30° + cos 60° × sin 30° =
(A) 0 (B) 1 (C) √3/2 (D) ½
Ans: (B) 1 — This is sin(60°+30°) = sin 90° = 1.
(A) 0 (B) 1 (C) √3/2 (D) ½
Ans: (B) 1 — This is sin(60°+30°) = sin 90° = 1.
MCQ 2 tan 45° + cot 45° =
(A) 0 (B) 1 (C) 2 (D) √2
Ans: (C) 2 — tan 45°=1; cot 45°=1. Sum=2.
(A) 0 (B) 1 (C) 2 (D) √2
Ans: (C) 2 — tan 45°=1; cot 45°=1. Sum=2.
A-R
A: sin(90°−A) = cos A.
R: Complementary angle identities relate trig ratios of complementary angles.
Ans: (A) Both correct, R explains A.
R: Complementary angle identities relate trig ratios of complementary angles.
Ans: (A) Both correct, R explains A.
📦 Case + SA + LA
Passage: A student notices that a right triangle has perpendicular = 5 and hypotenuse = 13. All six trig ratios are asked.
Q1 Find base. Ans: base = √(169−25) = √144 = 12.
Q2 Find all 6 ratios. Ans: sinA=5/13, cosA=12/13, tanA=5/12, cosecA=13/5, secA=13/12, cotA=12/5.
SA (2M) Evaluate: sin²30° + 2cos²45° + tan²60°.
Ans: (1/2)² + 2(1/√2)² + (√3)² = 1/4 + 1 + 3 = 4¼ = 17/4.
Ans: (1/2)² + 2(1/√2)² + (√3)² = 1/4 + 1 + 3 = 4¼ = 17/4.
LA (5M) Prove: (1 + secA)/secA = sin²A/(1−cosA).
Ans: RHS = sin²A/(1−cosA) = (1−cos²A)/(1−cosA) = (1−cosA)(1+cosA)/(1−cosA) = 1+cosA = 1 + 1/secA = (secA+1)/secA = LHS ✓
Ans: RHS = sin²A/(1−cosA) = (1−cos²A)/(1−cosA) = (1−cosA)(1+cosA)/(1−cosA) = 1+cosA = 1 + 1/secA = (secA+1)/secA = LHS ✓
🏔️ Applications of Trigonometry
Ch 9 · Heights and Distances · Angle of Elevation · Angle of Depression
📖 Key Concepts
Angle of Elevation
Angle between horizontal and the line of sight when looking up at an object. Observer is below object.
Angle of Depression
Angle between horizontal and line of sight when looking down. Observer is above object.
Approach
Draw a right triangle. Identify opposite, adjacent, hypotenuse relative to the given angle. Apply sin/cos/tan.
Alternate Angles
Angle of depression from top = angle of elevation from bottom (alternate interior angles — parallel lines).
🔑 Quick Reminders
Height = distance × tan(elevation angle)
Distance = height / tan(elevation angle)
tan 30° = 1/√3; tan 45° = 1; tan 60° = √3
✏️ Worked — Tower Height (Board 2023, 3M)
Q: A tower casts a shadow 40m long. Angle of elevation of sun is 30°. Find height of tower.
1
tan 30° = height / shadow = h / 40.
2
h = 40 × tan 30° = 40 × (1/√3) = 40/√3 = 40√3/3 ≈ 23.1 m.
✅ Height = 40√3/3 m
📝 MCQs + A-R
MCQ 1 A pole 6m high casts a shadow 6m long. The angle of elevation of the sun is:
(A) 30° (B) 45° (C) 60° (D) 90°
Ans: (B) 45° — tan θ = 6/6 = 1 → θ = 45°.
(A) 30° (B) 45° (C) 60° (D) 90°
Ans: (B) 45° — tan θ = 6/6 = 1 → θ = 45°.
A-R
A: Angle of depression = angle of elevation between same two points.
R: They are alternate interior angles formed by a transversal with parallel lines.
Ans: (A)
R: They are alternate interior angles formed by a transversal with parallel lines.
Ans: (A)
📦 Case + SA + LA
Passage: From the top of a 75m building, two cars are seen on opposite sides. Angles of depression are 60° and 45°.
Q1 Distance of car 1 (60° depression). Ans: d₁ = 75/tan60° = 75/√3 = 25√3 m.
Q2 Distance of car 2 (45° depression). Ans: d₂ = 75/tan45° = 75 m.
Q3 Distance between the two cars. Ans: 25√3 + 75 = 25(√3 + 3) m.
LA (5M) Two observers on opposite sides of a tower observe it at 30° and 60° elevation. Tower height = 100m. Find the distance between observers.
Ans: d₁ = 100/tan30° = 100√3; d₂ = 100/tan60° = 100/√3 = 100√3/3. Total = 100√3 + 100√3/3 = 400√3/3 ≈ 231 m.
Ans: d₁ = 100/tan30° = 100√3; d₂ = 100/tan60° = 100/√3 = 100√3/3. Total = 100√3 + 100√3/3 = 400√3/3 ≈ 231 m.
⭕ Circles
Ch 10 · Tangent · Chord · Theorems
📖 Key Concepts
Tangent
Line touching circle at exactly one point (point of tangency). Always perpendicular to radius at that point.
Tangents from External Point
Two tangents from same external point are equal in length. If PA and PB are tangents, PA = PB.
Angle in Semicircle
Angle subtended by diameter at any point on circle = 90°. Important theorem for geometry problems.
Number of Tangents
From external point: 2 tangents. From point on circle: 1 tangent. From point inside circle: 0 tangents.
🔑 Key Theorems
Radius ⊥ tangent at point of contact
Tangents from external point are equal
∠OPA = ∠OPB (tangents bisect the angle at external point)
✏️ Worked — Tangent Length (Board 2022, 2M)
Q: PA is tangent from P to circle (centre O, radius 5cm). OP = 13cm. Find PA.
1
∠OAP = 90° (radius ⊥ tangent).
2
PA = √(OP²−OA²) = √(169−25) = √144 = 12 cm.
✅ PA = 12 cm.
📝 MCQs + A-R + Case
MCQ 1 If OP = 10, OA = 6 (radius), then tangent length PA =
(A) 4 (B) 8 (C) 10 (D) 16
Ans: (B) 8 — √(100−36) = √64 = 8.
(A) 4 (B) 8 (C) 10 (D) 16
Ans: (B) 8 — √(100−36) = √64 = 8.
A-R
A: Two tangents from an external point make equal angles with the line joining that point to the centre.
R: The tangents are equal in length.
Ans: (B) Both true, but R does not directly explain A (the reason is OA=OB and OA⊥PA, OB⊥PB and OP common → congruent triangles → equal angles).
R: The tangents are equal in length.
Ans: (B) Both true, but R does not directly explain A (the reason is OA=OB and OA⊥PA, OB⊥PB and OP common → congruent triangles → equal angles).
Passage: A circle has centre O and radius 7cm. P is 25cm from centre. Two tangents PA and PB are drawn.
Q1 Find tangent length PA. Ans: √(625−49) = √576 = 24 cm.
Q2 PA = PB. Justify. Ans: Tangents from same external point are equal (theorem).
🥧 Areas Related to Circles
Ch 11 · Sector · Segment · Combinations
📖 Key Concepts
Area of Sector
Area = (θ/360) × πr². Arc length = (θ/360) × 2πr. θ is the angle at centre in degrees.
Area of Segment
Minor segment = Area of sector − Area of triangle. For standard angles use sin θ for triangle.
Perimeter of Sector
= 2r + arc length = 2r + (θ/360) × 2πr. Two radii + arc.
Common Combinations
Ring (annulus) = π(R²−r²). Flower petals: use sectors from square/equilateral triangle corners.
🔑 Key Formulae (use π = 22/7 unless told otherwise)
Area of sector = (θ/360) × πr²
Length of arc = (θ/360) × 2πr
Segment area = sector area − triangle area
Area of equil. △ = (√3/4)a² ; sin 60° = √3/2
✏️ Worked — Sector (Board 2023, 3M)
Q: Find area of sector with r = 14cm and θ = 45°.
1
Area = (45/360) × (22/7) × 14² = (1/8) × (22/7) × 196
2
= (1/8) × 22 × 28 = 616/8 = 77 cm².
✅ Area = 77 cm²
📝 MCQs + A-R + Case
MCQ 1 Area of sector with r=7cm, θ=90°:
(A) 38.5 cm² (B) 77 cm² (C) 154 cm² (D) 22 cm²
Ans: (A) 38.5 cm² — ¼ × 22/7 × 49 = 38.5.
(A) 38.5 cm² (B) 77 cm² (C) 154 cm² (D) 22 cm²
Ans: (A) 38.5 cm² — ¼ × 22/7 × 49 = 38.5.
A-R
A: Area of minor segment is less than area of major segment.
R: Minor segment corresponds to minor arc (angle < 180°).
Ans: (A)
R: Minor segment corresponds to minor arc (angle < 180°).
Ans: (A)
Passage: A brooch has a silver circle of diameter 28mm. Six sector-shaped petals each with angle 60° are cut out of it.
Q1 Area of each petal. Ans: r=14mm; (60/360)×(22/7)×196 = 102.67mm².
Q2 Total area of 6 petals. Ans: 6 × 102.67 = 616 mm² (= area of full circle! 6×60°=360°).
LA (5M) A horse is tied to one corner of a square field of side 15m with a 5m rope. Find area it can graze. Also find area if rope is 7m.
Ans: Corner angle = 90°. Area for 5m rope = (90/360)×π×25 = 25π/4 = 19.625 m². For 7m rope: 49π/4 = 38.5 m².
Ans: Corner angle = 90°. Area for 5m rope = (90/360)×π×25 = 25π/4 = 19.625 m². For 7m rope: 49π/4 = 38.5 m².
📦 Surface Areas and Volumes
Ch 12 · Cylinder · Cone · Sphere · Combinations · Conversion
📖 Key Concepts
Cylinder
CSA = 2πrh; TSA = 2πr(r+h); Volume = πr²h. Most common shape in board problems.
Cone
Slant height l = √(r²+h²); CSA = πrl; TSA = πr(l+r); Volume = ⅓πr²h.
Sphere & Hemisphere
Sphere: SA = 4πr²; V = (4/3)πr³. Hemisphere: CSA = 2πr²; TSA = 3πr²; V = (2/3)πr³.
Conversion of Shapes
When shape A melts to form shape B: Volume of A = Volume of B. Very common board question type.
🔑 All Formulae in One Place
Cylinder: V=πr²h; CSA=2πrh; TSA=2πr(r+h)
Cone: l=√(r²+h²); V=⅓πr²h; CSA=πrl; TSA=πr(r+l)
Sphere: V=(4/3)πr³; SA=4πr²
Hemisphere: V=(2/3)πr³; CSA=2πr²; TSA=3πr²
✏️ Worked — Conversion (Board 2023, 3M)
Q: A cone (r=5cm, h=12cm) is melted and recast as a sphere. Find radius of sphere.
1
V(cone) = ⅓ × π × 25 × 12 = 100π cm³.
2
(4/3)πR³ = 100π → R³ = 75 → R = ∛75 ≈ 4.22 cm.
✅ Radius ≈ 4.22 cm (or ∛75 cm exactly).
📝 MCQs + A-R + Case
MCQ 1 Volume of cone with r=3, h=7 (use π=22/7):
(A) 66 cm³ (B) 198 cm³ (C) 22 cm³ (D) 44 cm³
Ans: (A) 66 cm³ — ⅓×22/7×9×7=66.
(A) 66 cm³ (B) 198 cm³ (C) 22 cm³ (D) 44 cm³
Ans: (A) 66 cm³ — ⅓×22/7×9×7=66.
MCQ 2 TSA of a hemisphere of radius 7cm:
(A) 462 cm² (B) 308 cm² (C) 154 cm² (D) 616 cm²
Ans: (A) 462 cm² — 3πr² = 3×22/7×49 = 462.
(A) 462 cm² (B) 308 cm² (C) 154 cm² (D) 616 cm²
Ans: (A) 462 cm² — 3πr² = 3×22/7×49 = 462.
A-R
A: When a solid cone is melted into spheres, volume is conserved.
R: Mass is conserved; since density is same, volume is conserved.
Ans: (A)
R: Mass is conserved; since density is same, volume is conserved.
Ans: (A)
Passage: A toy is in the shape of a cone surmounted on a hemisphere, both of radius 3.5cm. Height of cone = 4cm.
Q1 Total volume? Ans: V(cone)+V(hemisphere) = ⅓π(3.5)²×4 + ⅔π(3.5)³ = 51.33+89.83 = 141.16 cm³.
Q2 Total surface area? Ans: CSA(cone)+CSA(hemisphere). l=√(3.5²+4²)=5.32. TSA = π×3.5×5.32 + 2π×3.5² = 58.5+77 = 135.5 cm².
📊 Statistics
Ch 13 · Mean · Median · Mode · Ogives
📖 Key Concepts
Mean (Grouped Data)
Direct Method: x̄ = Σfᵢxᵢ/Σfᵢ. Assumed Mean Method: x̄ = a + (Σfᵢdᵢ/Σfᵢ). xᵢ = class midpoint.
Median (Grouped)
Median = l + [(n/2−cf)/f] × h. l=lower class boundary; cf=cumulative freq before median class; f=freq of median class; h=class width.
Mode (Grouped)
Mode = l + [(f₁−f₀)/(2f₁−f₀−f₂)] × h. Modal class = class with highest frequency. f₁ = max freq.
Ogive
Cumulative frequency curve. Less-than ogive: plot (upper limit, cf). More-than ogive: plot (lower limit, more-than cf). Intersection point gives median.
🔑 Key Formulae
Mean: x̄ = Σfᵢxᵢ/Σfᵢ (direct method)
Median: l + [(n/2−cf)/f] × h
Mode: l + [(f₁−f₀)/(2f₁−f₀−f₂)] × h
Empirical: Mode = 3×Median − 2×Mean
✏️ Worked — Mode (Board 2023, 3M)
Q: Find mode for: Classes 10-20, 20-30, 30-40, 40-50 with frequencies 6, 11, 14, 5.
1
Modal class = 30-40 (highest freq = 14). l=30, f₁=14, f₀=11, f₂=5, h=10.
2
Mode = 30 + [(14-11)/(28-11-5)] × 10 = 30 + [3/12]×10 = 30 + 2.5 = 32.5.
✅ Mode = 32.5
📝 MCQs + A-R + Case
MCQ 1 Median of 3, 1, 4, 1, 5, 9, 2, 6 is:
(A) 3 (B) 3.5 (C) 4 (D) 5
Ans: (B) 3.5 — Sorted: 1,1,2,3,4,5,6,9. n=8; median=(3+4)/2=3.5.
(A) 3 (B) 3.5 (C) 4 (D) 5
Ans: (B) 3.5 — Sorted: 1,1,2,3,4,5,6,9. n=8; median=(3+4)/2=3.5.
A-R
A: Intersection of less-than and more-than ogives gives the median.
R: At the median, the cumulative frequency is n/2.
Ans: (A) Both true, R correctly explains A.
R: At the median, the cumulative frequency is n/2.
Ans: (A) Both true, R correctly explains A.
Passage: Monthly wages (₹) of 50 workers: 6000-7000: 5 workers; 7000-8000: 10; 8000-9000: 20; 9000-10000: 10; 10000-11000: 5.
Q1 Find mean wage. Ans: Midpoints: 6500,7500,8500,9500,10500. Σfx = 5×6500+10×7500+20×8500+10×9500+5×10500 = 32500+75000+170000+95000+52500=425000. Mean = 425000/50 = ₹8500.
Q2 Find modal class. Ans: 8000-9000 (highest freq = 20).
LA (5M) Find mean and median for data: Class 0-10: f=8; 10-20: f=16; 20-30: f=36; 30-40: f=34; 40-50: f=6.
Ans: n=100. Midpoints: 5,15,25,35,45. Σfx=8×5+16×15+36×25+34×35+6×45=40+240+900+1190+270=2640. Mean=26.4. Cumulative: 8,24,60,94,100. n/2=50; median class=20-30. Median=20+[(50-24)/36]×10=20+7.22=27.22.
Ans: n=100. Midpoints: 5,15,25,35,45. Σfx=8×5+16×15+36×25+34×35+6×45=40+240+900+1190+270=2640. Mean=26.4. Cumulative: 8,24,60,94,100. n/2=50; median class=20-30. Median=20+[(50-24)/36]×10=20+7.22=27.22.
🎲 Probability
Ch 14 · Classical Probability · Complementary Events · Cards & Dice
📖 Key Concepts
Classical Probability
P(E) = (Number of favourable outcomes) / (Total possible outcomes). Always between 0 and 1.
Complementary Event
P(not E) = 1 − P(E). P(E) + P(Ē) = 1 always.
Equally Likely Outcomes
A fair die has 6 equally likely outcomes. A fair coin has 2 equally likely outcomes. Standard deck: 52 equally likely cards.
Impossible & Sure Events
Impossible event: P = 0. Sure (certain) event: P = 1. All other probabilities are strictly between 0 and 1.
🔑 Common Probability Facts
P(E) = Favourable/Total; 0 ≤ P(E) ≤ 1
Standard deck: 52 cards; 4 suits; 13 per suit; 12 face cards; 4 aces
Die: sample space {1,2,3,4,5,6}; P(even) = 3/6 = ½
Two coins: sample space {HH, HT, TH, TT}; P(at least 1H) = 3/4
✏️ Worked — Card Problem (Board 2023, 2M)
Q: From a deck of 52 cards, find P(king) and P(not a face card).
1
Kings = 4. P(king) = 4/52 = 1/13.
2
Face cards = 12. P(not face) = 1 − 12/52 = 40/52 = 10/13.
✅ P(king) = 1/13; P(not face) = 10/13.
📝 MCQs + A-R + Case + SA + LA
MCQ 1 P(getting 7 on a die) =
(A) 1/6 (B) 7/6 (C) 0 (D) 1
Ans: (C) 0 — Impossible event; 7 is not on a standard die.
(A) 1/6 (B) 7/6 (C) 0 (D) 1
Ans: (C) 0 — Impossible event; 7 is not on a standard die.
MCQ 2 Two coins tossed. P(exactly one head) =
(A) ¼ (B) ½ (C) ¾ (D) 1
Ans: (B) ½ — Favourable: {HT, TH} = 2 out of 4.
(A) ¼ (B) ½ (C) ¾ (D) 1
Ans: (B) ½ — Favourable: {HT, TH} = 2 out of 4.
A-R
A: P(getting an even prime on a die) = 1/6.
R: 2 is the only even prime; P = 1/6.
Ans: (A) Both correct, R explains A.
R: 2 is the only even prime; P = 1/6.
Ans: (A) Both correct, R explains A.
Passage: A bag has 4 red, 3 blue, and 5 green balls. One ball is drawn at random.
Q1 P(red ball). Ans: 4/12 = 1/3.
Q2 P(not green ball). Ans: (4+3)/12 = 7/12.
Q3 P(blue or red). Ans: 7/12.
SA (2M) A box has 12 defective and 48 good items. P(picking defective) = ?
Ans: P = 12/60 = 1/5.
Ans: P = 12/60 = 1/5.
LA (5M) Cards numbered 1 to 30 are placed in a box. Find P(i) prime (ii) divisible by 5 (iii) perfect square (iv) a number > 25.
Ans: Total = 30. (i) Primes: 2,3,5,7,11,13,17,19,23,29 = 10 → P = 10/30 = 1/3. (ii) Div by 5: 5,10,15,20,25,30 = 6 → P = 6/30 = 1/5. (iii) Perfect sq: 1,4,9,16,25 = 5 → P = 5/30 = 1/6. (iv) >25: 26,27,28,29,30 = 5 → P = 5/30 = 1/6.
Ans: Total = 30. (i) Primes: 2,3,5,7,11,13,17,19,23,29 = 10 → P = 10/30 = 1/3. (ii) Div by 5: 5,10,15,20,25,30 = 6 → P = 6/30 = 1/5. (iii) Perfect sq: 1,4,9,16,25 = 5 → P = 5/30 = 1/6. (iv) >25: 26,27,28,29,30 = 5 → P = 5/30 = 1/6.
🧩 Solved Problems
Mixed Board-Style Problems — All Topics
✏️ Worked Examples
Problem 1 — AP and Sum (Board 2023, 4M)
Q: How many terms of AP: 3, 7, 11, 15... are needed so that sum = 406?
1
a = 3, d = 4. Sn = n/2 × [2×3 + (n-1)×4] = n/2 × [6 + 4n - 4] = n/2 × (4n + 2) = n(2n+1).
2
n(2n+1) = 406. 2n^2 + n - 406 = 0. Discriminant = 1 + 3248 = 3249 = 57^2. n = (-1 + 57)/4 = 14.
✅ Number of terms = 14.
📝 Practice Papers
Full-length 80-mark board-pattern papers · Mathematics Basic X
MATHEMATICS BASIC X — PRACTICE PAPER 1
Time: 3 Hours | Max Marks: 80 | CBSE 2024-25
SECTION A — MCQ (20×1=20)
Q1. HCF of 12 and 18 is: (a) 6 (b) 3 (c) 9 (d) 12 [Ans: a]
Q2. If p(x) = x² − 5x + 6, then p(2) = ? (a) 0 (b) −1 (c) 1 (d) 2 [Ans: a]
Q3. The discriminant of x² − 4x + 4 = 0 is: (a) 0 (b) 16 (c) −16 (d) 8 [Ans: a] [D=16−16=0, equal roots]
Q4. The 10th term of AP: 2, 7, 12, ... is: (a) 45 (b) 47 (c) 52 (d) 50 [Ans: b] [a+(n-1)d=2+9×5=47]
Q5. If ΔABC ~ ΔPQR and AB/PQ = 3/4, then area ratio = ? (a) 3:4 (b) 9:16 (c) 16:9 (d) 4:3 [Ans: b]
Q6. Distance between (0,0) and (3,4): (a) 3 (b) 4 (c) 5 (d) 7 [Ans: c]
Q7. sin²θ + cos²θ = ? (a) 0 (b) 1 (c) 2 (d) Depends on θ [Ans: b]
Q8. tan 45° = ? (a) 0 (b) 1/√2 (c) 1 (d) √3 [Ans: c]
Q9. Area of sector of angle 90° in circle of radius 7cm = ? (a) 38.5 cm² (b) 77 cm² (c) 154 cm² (d) 19.25 cm² [Ans: a] [90/360×π×49=38.5]
Q10. Curved surface area of cylinder, r=7, h=10: (a) 440 cm² (b) 880 cm² (c) 440π (d) 220π [Ans: a] [2×22/7×7×10=440]
Q11. Mean of 3,5,7,9,11 = ? (a) 5 (b) 7 (c) 9 (d) 6 [Ans: b]
Q12. P(getting head when tossing a fair coin) = ? (a) 1 (b) 0 (c) 1/2 (d) 1/4 [Ans: c]
Q13-20. [MCQs on Euclid's division lemma, Coordinate geometry (midpoint), Tangent properties, Volume of sphere, Median formula, Mode formula, Complementary events, Number of zeros of a polynomial] [8M]
SECTION B — VSA (5×2=10)
Q21. Find HCF(336, 54) using Euclid's algorithm. [2M]
Q22. Find the zeros of p(x) = x² − 3x − 4. [2M]
Q23. Find the roots of 2x² − 5x + 3 = 0 by factorisation. [2M]
Q24. In ΔABC, DE || BC, AD=2, DB=3, AE=4. Find AC. [2M]
Q25. The mean of 15 observations is 8. If 5 is added to each, find new mean. [2M]
SECTION C — SA (6×3=18)
Q26. Prove that √2 is irrational. [3M]
Q27. The sum of 3rd and 9th term of an AP is 24. If the 6th term is 12, find first term and common difference. [3M]
Q28. A ladder 10m long reaches a wall at point 8m high. Find distance of foot from wall. Draw diagram. [3M]
Q29. From top of a 20m tower, the angle of depression of a car is 30°. Find distance of car from base. [3M]
Q30. Find the area of shaded region: circle of radius 14cm with square of side 10cm inscribed. [3M]
Q31. Find mean of following data using assumed mean method: [3M]
Class: 0-10, 10-20, 20-30, 30-40, 40-50; Frequency: 5, 10, 15, 12, 8
SECTION D — LA (4×5=20) + Case (2×4=8) + Source
Q32. Solve graphically: 2x + y = 8; x + y = 5. Find area of triangle formed with x-axis. [5M]
Q33. Sides of a right triangle are x, x+1, x+2 (hypotenuse = x+2). Find sides. Also find the perimeter. [5M]
Q34. A hemispherical bowl of radius 9cm is full of ice cream. It is distributed in cones (r=3cm, h=12cm) with hemispherical top. How many students get ice cream? [5M]
Q35. (a) A bag has 3 red, 5 white, 7 black balls. Ball drawn randomly. P(red) = ? P(not black) = ? P(white or red) = ? (b) Two dice thrown. P(sum = 8)? P(same number)? [5M]
Q36. Case: Rohan's teacher draws a circle on the board. Tangents from external point P touch at A and B. PA=PB=5cm, OA=3cm. Answer: (i) OAP angle (ii) OP length (iii) Area of OAPB (iv) PA/OA. [4M]
Q37. Case: Ananya collects data on daily screen time of 30 students. Modal class is 4-5hrs, frequency = 12; preceding class frequency = 8; following class = 6. Find mode. [4M]
MATHEMATICS BASIC X — PRACTICE PAPER 2
Different chapter focus
Q1-37. Full 80-mark paper focusing on: LCM method problems, Relationship between zeros and coefficients, Sum of AP terms, Coordinate geometry (distance and section formula), Trigonometric identities, Heights and distances (two angle problems), Surface area combination shapes, Median using ogive. [80M]
MATHEMATICS BASIC X — PRACTICE PAPER 3 (Challenge)
Slightly harder — for extra practice
Q1. Prove: If n is odd, n² is also odd. [2M]
Q2. If zeros of x² + px + q are double of zeros of x² + 3x + 2, find p and q. [4M]
Q3. A train travels 360km at uniform speed. If speed were 5km/h more, it would take 1hr less. Find speed. [5M]
Q4. AP: sum of first 7 terms = 49, sum of first 17 terms = 289. Find sum of n terms. [5M]
Q5. ΔABC, right-angled at C. If tanA = 5/12, find all other trigonometric ratios of A. [4M]
Q6. Prove: (sin A + cos A)² + (sin A − cos A)² = 2. [3M]
Q7. A sphere of diameter 12cm is dropped in a cylinder of radius 6cm containing water. How much does water level rise? [5M]
📜 PYQ Bank — Mathematics Basic (Class 10)
CBSE board questions 2019–2024 — Real Numbers, Polynomials, Algebra, Geometry, Trigonometry, Statistics
⚠️ Maths Basic — same syllabus as Standard, slightly easier application questions
Real Numbers — HCF/LCM (2022, 2023, 2024)
Q: Find HCF of 96 and 404 by prime factorisation. (2 marks)
Answer: 96 = 2⁵ × 3; 404 = 2² × 101; HCF = 2² = 4.
Also asked: Prove √3 irrational (proof by contradiction), LCM × HCF = product of two numbers (always verify).
Answer: 96 = 2⁵ × 3; 404 = 2² × 101; HCF = 2² = 4.
Also asked: Prove √3 irrational (proof by contradiction), LCM × HCF = product of two numbers (always verify).
Quadratic Equations — Factorisation (2021, 2022, 2023)
Q: Solve: x² + 5x + 6 = 0 by factorisation. (2 marks)
Answer: (x + 2)(x + 3) = 0 → x = −2 or x = −3.
Also asked: Complete the square method, quadratic formula, nature of roots (b²−4ac).
Answer: (x + 2)(x + 3) = 0 → x = −2 or x = −3.
Also asked: Complete the square method, quadratic formula, nature of roots (b²−4ac).
Triangles — Similarity & Pythagoras (2020, 2021, 2023)
Q: In △ABC, DE ∥ BC. If AD = 3cm, DB = 4cm, AE = 2.4cm, find EC. (2 marks)
Answer: By BPT: AD/DB = AE/EC → 3/4 = 2.4/EC → EC = 3.2 cm.
Also asked: Pythagoras theorem proof, similar triangle area ratio.
Answer: By BPT: AD/DB = AE/EC → 3/4 = 2.4/EC → EC = 3.2 cm.
Also asked: Pythagoras theorem proof, similar triangle area ratio.
Coordinate Geometry — Distance & Section Formula (2021, 2022)
Q: Find the distance between A(3, 4) and B(−2, −8). (2 marks)
Answer: d = √[(3−(−2))² + (4−(−8))²] = √[25 + 144] = √169 = 13 units.
Also: Midpoint, section formula, area of triangle using coordinates.
Answer: d = √[(3−(−2))² + (4−(−8))²] = √[25 + 144] = √169 = 13 units.
Also: Midpoint, section formula, area of triangle using coordinates.
Trigonometry — Identities (2019–2024 every year)
Q: Prove: (sin A + cosec A)² + (cos A + sec A)² = 7 + tan²A + cot²A. (3 marks)
Strategy: Expand LHS, use sin²+cos²=1, 1+tan²=sec², 1+cot²=cosec². Always expand the more complex side.
Also asked: Values of sin/cos/tan 30°, 45°, 60° applied in equations.
Strategy: Expand LHS, use sin²+cos²=1, 1+tan²=sec², 1+cot²=cosec². Always expand the more complex side.
Also asked: Values of sin/cos/tan 30°, 45°, 60° applied in equations.
Statistics — Mean/Median/Mode (2020, 2021, 2022, 2023)
Q: Find the mode of: 3, 5, 7, 5, 8, 3, 5, 7, 5 → Mode = 5 (appears 4 times). (1 mark)
3-mark Q: Find mean using assumed mean method from frequency table. Set up: d = x − A, Σfd, apply formula x̄ = A + Σfd/Σf.
5-mark Q: Median from cumulative frequency ogive — plot and find n/2 point.
3-mark Q: Find mean using assumed mean method from frequency table. Set up: d = x − A, Σfd, apply formula x̄ = A + Σfd/Σf.
5-mark Q: Median from cumulative frequency ogive — plot and find n/2 point.
Surface Area & Volume (2020, 2022, 2024)
Q: A cone of radius 6 cm and slant height 10 cm. Find total surface area. (3 marks)
Answer: TSA = πrl + πr² = π(6)(10) + π(36) = 60π + 36π = 96π = 301.44 cm².
Also asked: Cylinder + hemisphere combination, volume of frustum, conversion questions.
Answer: TSA = πrl + πr² = π(6)(10) + π(36) = 60π + 36π = 96π = 301.44 cm².
Also asked: Cylinder + hemisphere combination, volume of frustum, conversion questions.
Probability (2019–2024 every year)
Q: A card is drawn from a deck of 52 cards. Find probability of getting a king. (1 mark)
Answer: P = 4/52 = 1/13.
3-mark Q (2023): Two dice rolled — find P(sum = 7). Sample space = 36; favourable = (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) = 6; P = 6/36 = 1/6.
Answer: P = 4/52 = 1/13.
3-mark Q (2023): Two dice rolled — find P(sum = 7). Sample space = 36; favourable = (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) = 6; P = 6/36 = 1/6.
📋 Maths Basic Marks Pattern: MCQ+Assertion (20×1=20) + SA-I (5×2=10) + SA-II (6×3=18) + LA (4×5=20) + Case-based (3×4=12) = 80 marks. Trigonometry + Statistics + Algebra = 40% of paper. Master these 3.
❌ Mistake Bank — Mathematics Basic
Most costly errors in Maths Basic boards — these lose marks needlessly
Trigonometry — Identity Application Error
Mistake: Writing sin²A + cos²A = 0 instead of 1 under exam pressure — cancels all subsequent steps.
Fix: Write the three Pythagorean identities at the top of every trig proof as a reference before starting. They are: sin²+cos²=1, 1+tan²=sec², 1+cot²=cosec².
Fix: Write the three Pythagorean identities at the top of every trig proof as a reference before starting. They are: sin²+cos²=1, 1+tan²=sec², 1+cot²=cosec².
Quadratic Formula — Sign Error in Discriminant
Mistake: Computing b²−4ac as (−5)²−4(1)(6) = 25−24 = 1 but then writing x = −b ± √1 / 2a with wrong sign of b.
Fix: Write the formula explicitly: x = [−(−5) ± √1] / 2 → [5 ± 1]/2 = 3 or 2. Box the formula first, substitute second.
Fix: Write the formula explicitly: x = [−(−5) ± √1] / 2 → [5 ± 1]/2 = 3 or 2. Box the formula first, substitute second.
Statistics — Wrong Class Mark
Mistake: For class 10–20, using class mark = 10 instead of 15 (lower + upper limit divided by 2).
Fix: Class mark = (lower + upper) / 2 always. For 10–20 → (10+20)/2 = 15. This error affects the entire mean calculation.
Fix: Class mark = (lower + upper) / 2 always. For 10–20 → (10+20)/2 = 15. This error affects the entire mean calculation.
Probability — Not Simplifying Fraction
Mistake: Writing P = 4/52 without simplifying to 1/13 — CBSE marking scheme shows simplified form; unsimplified may lose ½ mark.
Fix: Always simplify probability fractions. Check HCF of numerator and denominator before writing final answer.
Fix: Always simplify probability fractions. Check HCF of numerator and denominator before writing final answer.
BPT — Mixing AD/DB with AB/DB
Mistake: In Basic Proportionality Theorem, writing AD/AB = AE/AC instead of AD/DB = AE/EC.
Fix: BPT states: if DE ∥ BC then AD/DB = AE/EC (parts, not whole). Draw the diagram and label before applying formula.
Fix: BPT states: if DE ∥ BC then AD/DB = AE/EC (parts, not whole). Draw the diagram and label before applying formula.
Circles — Tangent Length Formula
Mistake: Forgetting that tangent from external point: PA² = PC × PB (power of a point) or PA = PB (equal tangents from same external point).
Fix: Equal tangents from external point is used in 90% of CBSE circle tangent problems. Draw the figure — PT² = PO² − r².
Fix: Equal tangents from external point is used in 90% of CBSE circle tangent problems. Draw the figure — PT² = PO² − r².
📓 Maths Basic Rule: For every 3-mark or 5-mark question, write: "Given → To Prove/Find → Construction (if needed) → Proof/Solution → Result." This structured approach earns step marks even when the final answer is wrong.
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👁️ Examiner's Eye — Mathematics Basic
What the CBSE Maths Basic examiner checks — how to earn step marks
Step Marks — How They Work
3-mark question: 1 mark for correct method/formula + 1 mark for substitution + 1 mark for answer.
5-mark question: 1 mark formula + 2 marks working + 1 mark correct steps + 1 mark final answer.
If final answer is wrong but all steps are correct → 4/5 marks. Never leave a question blank.
5-mark question: 1 mark formula + 2 marks working + 1 mark correct steps + 1 mark final answer.
If final answer is wrong but all steps are correct → 4/5 marks. Never leave a question blank.
Proof Questions — What Examiner Checks
Mandatory: "Given", "To Prove", and "Proof" headings. Missing "Given" or "To Prove" → −1 mark. The proof must use valid mathematical steps — circular proofs (assuming what you need to prove) get zero.
Construction Questions
Examiner checks: Correct arc marks visible (don't erase them), angles within ±1°, lengths within ±1mm. Write steps of construction below the figure. Each step = ½ to 1 mark. Pencil + compass only — no freehand allowed.
Case-Based Questions (4 marks each)
Pattern: A real-life scenario + 4 sub-questions (1 mark each). Read the scenario once, underline key measurements. The 4 questions are usually: formula identification, substitution, calculation, interpretation.
These are the easiest marks in the paper — don't skip them.
These are the easiest marks in the paper — don't skip them.
Diagram and Labelling
Mandatory for: All geometry proofs, trigonometry height/distance, surface area problems. Unlabelled or missing diagram = −1 mark in 5-mark questions. Draw neatly but quickly — 30 seconds per diagram.
MCQ and Assertion-Reason
No step marks here — right or wrong. Common trap in Assertion-Reason: both A and R are true but R does NOT explain A → answer is (b), not (a). For each A-R question, check: (1) Is A true? (2) Is R true? (3) If both true, does R justify A? Answer logically.
🎯 Maths Basic Examiner Secret: Write units in every answer — cm², m³, degrees. "96π cm²" not just "96π". Missing units can cost ½ mark per question. In a 5-mark question with 4 sub-answers, 4 missing units = 2 marks lost.
🔥 High-Probability — Mathematics Basic
Topics that appear in Maths Basic board every year — don't miss these
📌 Probability — 100% probability
Every year: 1-mark card probability, 3-mark two-dice sum problem, case-based probability.
Must know: P(event) = favourable outcomes / total outcomes. Cards: 52 total, 4 suits of 13 each. Dice: 36 outcomes for two dice. These are calculation-free if sample space is memorised.
Must know: P(event) = favourable outcomes / total outcomes. Cards: 52 total, 4 suits of 13 each. Dice: 36 outcomes for two dice. These are calculation-free if sample space is memorised.
📌 Statistics (Mean/Median/Mode) — 100% probability
Every year: Find mean using assumed mean method (3 marks), mode from frequency table (1 mark), median using ogive (5 marks).
Formulas: Mean = A + Σfd/Σf; Mode = l + [(f₁−f₀)/(2f₁−f₀−f₂)]×h; Median = l + [(n/2−cf)/f]×h
Formulas: Mean = A + Σfd/Σf; Mode = l + [(f₁−f₀)/(2f₁−f₀−f₂)]×h; Median = l + [(n/2−cf)/f]×h
📌 Quadratic Equations — 92% probability
Every year: Solve by factorisation (2 marks) + word problem (5 marks).
Word problems: Train speed, age problems, rectangles. Identify: let x = ?, form equation, solve by factorisation or formula, reject negative root if needed.
Word problems: Train speed, age problems, rectangles. Identify: let x = ?, form equation, solve by factorisation or formula, reject negative root if needed.
📌 Trigonometric Identities — 90% probability
Every year: Prove identity (3 marks) + evaluate expression using standard angles (1-2 marks).
Must know: sin 30°=½, cos 30°=√3/2, tan 45°=1, sin 90°=1, cos 0°=1. Memorise 5 identities and use them in proofs.
Must know: sin 30°=½, cos 30°=√3/2, tan 45°=1, sin 90°=1, cos 0°=1. Memorise 5 identities and use them in proofs.
📌 Surface Area & Volume — 88% probability
Every year: Cylinder + hemisphere or cone + cylinder combined volume (5 marks).
Formula list: CSA cylinder = 2πrh; CSA cone = πrl; SA sphere = 4πr²; Volume: πr²h (cyl), ⅓πr²h (cone), ⁴⁄₃πr³ (sphere).
Formula list: CSA cylinder = 2πrh; CSA cone = πrl; SA sphere = 4πr²; Volume: πr²h (cyl), ⅓πr²h (cone), ⁴⁄₃πr³ (sphere).
📌 Similar Triangles / BPT — 85% probability
Every year: BPT application (2 marks) + similarity condition proof (3-5 marks).
Must know: AA, SSS, SAS similarity criteria. Ratio of areas = (ratio of sides)². Pythagoras theorem for right triangles.
Must know: AA, SSS, SAS similarity criteria. Ratio of areas = (ratio of sides)². Pythagoras theorem for right triangles.
📌 Arithmetic Progressions — 90% probability
Every year: Find nth term (1-2 marks) + sum of n terms (3 marks) + word problem on AP (5 marks).
Formulas: aₙ = a + (n−1)d; Sₙ = n/2 × [2a + (n−1)d] or n/2 × (a + l). Always identify a, d, n first.
Formulas: aₙ = a + (n−1)d; Sₙ = n/2 × [2a + (n−1)d] or n/2 × (a + l). Always identify a, d, n first.
📌 HCF/LCM + Proof of Irrationality — 88% probability
Every year: HCF by prime factorisation (2 marks) + prove √2 or √3 irrational (3 marks).
Proof structure: Assume √2 = p/q (rational, HCF(p,q)=1) → 2 divides p → 2 divides p → 2 divides q → contradiction → irrational. This exact proof appears 4/6 years.
Proof structure: Assume √2 = p/q (rational, HCF(p,q)=1) → 2 divides p → 2 divides p → 2 divides q → contradiction → irrational. This exact proof appears 4/6 years.
🎯 Maths Basic Top-5 (by marks): (1) Statistics 5-mark Ogive = 5 marks (2) Surface Area/Volume = 5 marks (3) Quadratic word problem = 5 marks (4) Similar Triangles proof = 5 marks (5) Probability 3-mark = 3 marks. These 5 = 23 marks. Get these right.
⏱️ Last-Minute Revision — Mathematics Basic
48 hours before Maths Basic board — formula recall and strategy
📐 Trigonometry Table
Write this from memory right now:
sin 0°=0, sin 30°=½, sin 45°=1/√2, sin 60°=√3/2, sin 90°=1
cos is reverse of sin (cos 0°=1, cos 90°=0)
tan = sin/cos: tan 30°=1/√3, tan 45°=1, tan 60°=√3
Identities: sin²θ+cos²θ=1 | sec²θ=1+tan²θ | cosec²θ=1+cot²θ
sin 0°=0, sin 30°=½, sin 45°=1/√2, sin 60°=√3/2, sin 90°=1
cos is reverse of sin (cos 0°=1, cos 90°=0)
tan = sin/cos: tan 30°=1/√3, tan 45°=1, tan 60°=√3
Identities: sin²θ+cos²θ=1 | sec²θ=1+tan²θ | cosec²θ=1+cot²θ
📊 Statistics Formulas
Mean (Assumed Mean): x̄ = A + Σfd/Σf where d = x − A
Mode: l + [(f₁−f₀)/(2f₁−f₀−f₂)] × h
Median: l + [(n/2 − cf) / f] × h
For ogive: Plot (upper boundary, cumulative frequency); find n/2 on y-axis; draw horizontal line to curve; drop vertical to x-axis = median.
Mode: l + [(f₁−f₀)/(2f₁−f₀−f₂)] × h
Median: l + [(n/2 − cf) / f] × h
For ogive: Plot (upper boundary, cumulative frequency); find n/2 on y-axis; draw horizontal line to curve; drop vertical to x-axis = median.
📏 Area and Volume Quick List
Write these 8 formulas from memory:
Circle: A = πr², C = 2πr
Cylinder: V = πr²h, CSA = 2πrh
Cone: V = ⅓πr²h, CSA = πrl, l = √(r²+h²)
Sphere: V = ⁴⁄₃πr³, SA = 4πr²
Hemisphere: V = ⅔πr³, CSA = 2πr²
Circle: A = πr², C = 2πr
Cylinder: V = πr²h, CSA = 2πrh
Cone: V = ⅓πr²h, CSA = πrl, l = √(r²+h²)
Sphere: V = ⁴⁄₃πr³, SA = 4πr²
Hemisphere: V = ⅔πr³, CSA = 2πr²
🎲 Probability Quick-Recall
Card probability: 52 cards; 4 suits (Hearts, Diamonds, Clubs, Spades); 13 per suit; 4 kings, 4 queens, 4 aces, 4 jacks; 26 red, 26 black.
Dice: 1 die = 6 outcomes; 2 dice = 36 outcomes. P(sum=7) = 6/36 = 1/6. P(sum=12) = 1/36.
Bag problems: P = favourable/total. Always simplify.
Dice: 1 die = 6 outcomes; 2 dice = 36 outcomes. P(sum=7) = 6/36 = 1/6. P(sum=12) = 1/36.
Bag problems: P = favourable/total. Always simplify.
✅ Maths Basic Checklist
✅ Trigonometry table written from memory
✅ 3 statistics formulas (mean/mode/median) written
✅ 8 area/volume formulas memorised
✅ Irrational number proof written (√2 or √3)
✅ BPT theorem and application practised
✅ Quadratic factorisation — 5 examples
✅ AP nth term + sum formulas written
✅ 1 full mock paper timed in 3 hours
✅ 3 statistics formulas (mean/mode/median) written
✅ 8 area/volume formulas memorised
✅ Irrational number proof written (√2 or √3)
✅ BPT theorem and application practised
✅ Quadratic factorisation — 5 examples
✅ AP nth term + sum formulas written
✅ 1 full mock paper timed in 3 hours
⏰ Exam Time Plan (3 hours, 80 marks)
Section-wise time:
Section A MCQ (20 marks): 25 minutes
Section B SA-I 2-mark (10 marks): 15 minutes
Section C SA-II 3-mark (18 marks): 30 minutes
Section D LA 5-mark (20 marks): 40 minutes
Section E Case-based (12 marks): 25 minutes
Check + fill blanks: 5 minutes
Section A MCQ (20 marks): 25 minutes
Section B SA-I 2-mark (10 marks): 15 minutes
Section C SA-II 3-mark (18 marks): 30 minutes
Section D LA 5-mark (20 marks): 40 minutes
Section E Case-based (12 marks): 25 minutes
Check + fill blanks: 5 minutes
⏱ Maths Basic Last Tip: Attempt Section E (Case-based, 12 marks) right after MCQs — these are diagram-based, not hard, and worth 15% of paper. Students who do Section E early score 10–12/12. Students who leave it for last often score 4–6/12 due to time pressure.
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