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Class 9 Maths important questions chapter-wise 2026-27, with marks

CMDLearn · 4 November 2026

These are 28 important questions for Class 9 Maths in 2026-27, two from each of the 14 chapters of the new NCERT books, Ganita Manjari Part I (Chapters 1 to 8) and Part II (Chapters 9 to 14). Each question carries a marks label (1, 2, 3 or 5) and a worked answer, and each chapter heading gives the book's printed pages.

The questions are our own. None is a question from the books, and none copies the numbers of our chapter guides, so you can use them as fresh practice after the chapter. A teacher has not yet reviewed them.

How the 80 marks are split by unit

CBSE's Class IX Mathematics curriculum for 2026-27 gives the theory paper 80 marks, in six units (course structure, page 4):

Two things to know about this table:

• The CBSE list under Geometry also names "Introduction to Euclid's Geometry", "Lines and Angles" and "Triangles – Congruence Theorems". These are not chapter titles in Part I or Part II as published (we checked both contents pages on 4 November 2026). Ask your teacher what your school's paper covers. Chapter 11, The World of Algorithms, is in the book but not in CBSE's unit list.
• The chapter column is our reading of the two lists side by side. CBSE gives marks only per unit, never per chapter.

The paper is 3 hours. CBSE sets its mix by type of thinking: 43 marks (about 54%) for remembering and understanding, 19 marks (24%) for applying, and 18 marks (22%) for analysing, evaluating and creating (question paper design, page 13). Another 20 marks are internal assessment: pen-paper tests and multiple assessment (10), portfolio (5) and lab practical (5) (page 13).

Class 9 has no board exam: your school sets the half-yearly and annual papers. So the 1, 2, 3 and 5-mark labels below are our guide to how much working each question needs, not CBSE's labels.

The questions, chapter by chapter

1. Orienting Yourself: The Use of Coordinates (Part I, Chapter 1, printed pages 1 to 15)

Q1 (1 mark, multiple choice). The point (0, −8) lies: (A) in Quadrant III (B) in Quadrant IV (C) on the y-axis (D) on the x-axis

Answer: (C). Its x-coordinate is 0, so the point is on the y-axis, 8 units below the origin. A point on an axis is in no quadrant.

Q2 (3 marks, short answer). A(−1, −2), B(11, −2) and C(11, 3) are three corners of a triangular garden on a grid. Find AB, BC and AC, and show that the triangle has a right angle. At which corner is it?

Answer: AB = 12, BC = 5, AC = 13 units; the right angle is at B.

• A and B have the same y-coordinate, so AB = 11 − (−1) = 12. B and C have the same x-coordinate, so BC = 3 − (−2) = 5.
• AC = √((11 + 1)² + (3 + 2)²) = √(144 + 25) = √169 = 13.
• 12² + 5² = 144 + 25 = 169 = 13², so the angle opposite AC, the angle at B, is a right angle (the distance between two points comes from the Baudhāyana–Pythagoras theorem, page 8).

2. Introduction to Linear Polynomials (Part I, Chapter 2, printed pages 16 to 40)

Q1 (1 mark, multiple choice). Which of these is a linear polynomial in x? (A) 3x² + 2 (B) 5 − 4x (C) 1/x + 1 (D) x³

Answer: (B). 5 − 4x has degree 1. (A) has degree 2, (D) degree 3, and (C) is not a polynomial, because x is in the denominator.

Q2 (3 marks, short answer). A cab company charges a fixed ₹45 for a ride plus ₹13 for every kilometre. (a) Write the fare f(d), in rupees, for a ride of d km. (b) Find the fare for 11 km. (c) A ride cost ₹253. How long was it?

Answer: (a) f(d) = 13d + 45; (b) ₹188; (c) 16 km.

• (a) Every kilometre adds the same ₹13, so the fare is a linear polynomial in d with the fixed charge as its constant term: f(d) = 13d + 45.
• (b) f(11) = 143 + 45 = ₹188.
• (c) 13d + 45 = 253 gives 13d = 208, so d = 16 km.

3. The World of Numbers (Part I, Chapter 3, printed pages 41 to 67)

Q1 (1 mark, assertion and reason). Assertion (A): 11/16 has a terminating decimal expansion. Reason (R): In lowest terms, the denominator 16 has no prime factor other than 2. (A) Both A and R are true, and R explains A. (B) Both are true, but R does not explain A. (C) A is true, R is false. (D) A is false, R is true.

Answer: (A). 11/16 = 0.6875, which terminates. A fraction p/q in lowest terms terminates exactly when q's prime factors are only 2s, only 5s, or 2s and 5s (page 58), and 16 = 2⁴.

Q2 (3 marks, short answer). (a) Write 0.363636… (the block 36 repeating) as a fraction p/q in lowest terms. (b) Find one rational number between this number and 5/11.

Answer: (a) 4/11; (b) for example 9/22.

• (a) Let x = 0.3636…. Then 100x = 36.3636…, and subtracting, 99x = 36, so x = 36/99 = 4/11.
• (b) The average of 4/11 and 5/11 is (8/22 + 10/22) ÷ 2 = 9/22, and 8/22 < 9/22 < 10/22. (Any rational number strictly between them is correct.)

4. Exploring Algebraic Identities (Part I, Chapter 4, printed pages 68 to 91)

Q1 (2 marks, very short answer). Use an identity to find 103 × 97 without long multiplication.

Answer: 9991. 103 × 97 = (100 + 3)(100 − 3) = 100² − 3² = 10000 − 9 = 9991, using (a + b)(a − b) = a² − b² (page 77).

Q2 (3 marks, short answer). Factorise 4x² + 12xy + 9y². Then use your factorised form to find its value when x = 2 and y = −1, and check by putting the values into the original expression.

Answer: (2x + 3y)²; value 1.

• 4x² = (2x)², 9y² = (3y)² and 12xy = 2 × (2x) × (3y), so the expression is (2x + 3y)², by (a + b)² = a² + 2ab + b² (page 70).
• At x = 2, y = −1: (4 − 3)² = 1² = 1.
• Check: 4(4) + 12(2)(−1) + 9(1) = 16 − 24 + 9 = 1 ✓.

5. Circles, the chapter "Up and Down, and Round and Round" (Part I, Chapter 5, printed pages 92 to 117)

Q1 (2 marks, very short answer). A chord of a circle is 30 cm long, and its distance from the centre is 8 cm. Find the radius of the circle.

Answer: 17 cm. A line dropped at right angles from the centre onto a chord cuts the chord into two equal halves (Theorem 5, page 102), so it meets the chord 15 cm from each end. The radius is the hypotenuse of a right triangle with legs 15 cm and 8 cm: √(225 + 64) = √289 = 17 cm.

Q2 (3 marks, short answer). A, B, C and D are points on a circle with centre O. The arc AB that does not contain C subtends ∠AOB = 136° at the centre. C is on the major arc AB and D is on the minor arc AB. Find ∠ACB and ∠ADB.

Answer: ∠ACB = 68°, ∠ADB = 112°.

• The angle an arc subtends at the centre is double the angle it subtends at a point on the rest of the circle (page 108), so ∠ACB = 136° ÷ 2 = 68°.
• ACBD is a cyclic quadrilateral, and its opposite angles add up to 180° (pages 112 to 113), so ∠ADB = 180° − 68° = 112°.

6. Measuring Space: Perimeter and Area (Part I, Chapter 6, printed pages 118 to 154)

Q1 (3 marks, short answer). A triangular plot has sides 9 m, 10 m and 17 m. Find its area using Heron's formula.

Answer: 36 m².

• Semi-perimeter s = (9 + 10 + 17) ÷ 2 = 18 (page 134).
• Heron's formula multiplies s by its three differences from the sides: area = √(18 × (18 − 9) × (18 − 10) × (18 − 17)) = √(18 × 9 × 8 × 1) = √1296 = 36 m².

Q2 (3 marks, short answer). A sector of a circle has radius 28 cm and angle 45° at the centre. Taking π = 22/7, find (a) the length of its arc, (b) its perimeter, (c) its area.

Answer: (a) 22 cm; (b) 78 cm; (c) 308 cm².

• (a) The arc is 45/360 = 1/8 of the circumference: (1/8) × 2 × (22/7) × 28 = (1/8) × 176 = 22 cm (length of an arc, page 126).
• (b) Perimeter = arc + two radii = 22 + 28 + 28 = 78 cm.
• (c) Area = (1/8) × (22/7) × 28 × 28 = (1/8) × 2464 = 308 cm² (area of a sector, pages 147 to 148).

7. The Mathematics of Maybe: Introduction to Probability (Part I, Chapter 7, printed pages 155 to 173)

Q1 (1 mark, multiple choice). Which of these cannot be the probability of an event? (A) 0 (B) 0.35 (C) 1.25 (D) 1

Answer: (C). A probability is always between 0 and 1, both included (page 157). 1.25 is greater than 1.

Q2 (3 marks, short answer). A spinner has three equal sectors marked 1, 2 and 3. It is spun twice. (a) Draw a tree diagram, or list the sample space. How many outcomes are there? (b) Find the probability that the two numbers add up to 4. (c) Find the probability that both numbers are odd.

Answer: (a) 9 outcomes; (b) 1/3; (c) 4/9.

• (a) Each spin has 3 equally likely results, so there are 3 × 3 = 9 outcomes: (1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3) (sample spaces and tree diagrams, pages 166 to 169).
• (b) Sum 4: (1,3), (2,2), (3,1), so 3/9 = 1/3.
• (c) Both odd: (1,1), (1,3), (3,1), (3,3), so 4/9.

8. Sequences and progressions, the chapter "Predicting What Comes Next" (Part I, Chapter 8, printed pages 174 to 196)

Q1 (2 marks, very short answer). A sequence has the recursive rule a₁ = 4 and aₙ₊₁ = 3aₙ − 2. (a) Write its first four terms. (b) Is it an arithmetic progression, a geometric progression, or neither?

Answer: (a) 4, 10, 28, 82; (b) neither.

• (a) a₂ = 12 − 2 = 10, a₃ = 30 − 2 = 28, a₄ = 84 − 2 = 82 (recursive rules, page 178).
• (b) The differences are 6, 18, 54, not equal, so it is not an AP. The ratios are 10/4 = 2.5 and 28/10 = 2.8, not equal, so it is not a GP. (An explicit rule that fits is aₙ = 3ⁿ + 1.)

Q2 (3 marks, short answer). (a) Find the sum of the first 80 natural numbers. (b) Use it to find 41 + 42 + 43 + … + 80.

Answer: (a) 3240; (b) 2420.

• (a) 1 + 2 + … + n = n(n + 1)/2 (page 184), so 80 × 81 ÷ 2 = 3240.
• (b) Subtract the first 40: 40 × 41 ÷ 2 = 820, and 3240 − 820 = 2420.

9. Propositions and their Converses (Part II, Chapter 9, printed pages 1 to 7)

Q1 (1 mark, very short answer). Write the converse of "If a whole number is divisible by 6, then it is divisible by 3." Is the converse true?

Answer: The converse is "If a whole number is divisible by 3, then it is divisible by 6." It is false: 9 is divisible by 3 but not by 6. (A statement and its converse need not both be true, page 1.)

Q2 (2 marks, very short answer). "If x² = 49, then x = 7." Is this proposition true? Write its converse and decide whether the converse is true. Give a reason or a counterexample for each.

Answer: The proposition is false: x = −7 also gives x² = 49, so x² = 49 does not force x = 7. Its converse, "If x = 7, then x² = 49", is true, because 7 × 7 = 49. So a false proposition can have a true converse.

10. How Quantities Combine: Understanding Data (Part II, Chapter 10, printed pages 8 to 36)

Q1 (3 marks, short answer). In a test, the 25 students of Section A scored a mean of 58 marks, and the 35 students of Section B scored a mean of 70 marks. Find the mean of all 60 students. Why is the answer not 64?

Answer: 65.

• Total marks: 25 × 58 = 1450 and 35 × 70 = 2450, together 3900. Mean = 3900 ÷ 60 = 65.
• 64 is the plain average of 58 and 70. It would be right only if both sections had the same number of students. Section B is larger, so the combined mean is pulled towards 70: it is a weighted average (pages 8 and 12).

Q2 (2 marks, very short answer). 4 litres of a drink that is 15% juice are mixed with 6 litres of a drink that is 40% juice. What percentage of the mixture is juice?

Answer: 30%. Juice: 15% of 4 L = 0.6 L, and 40% of 6 L = 2.4 L, so 3 L of juice in 10 L, which is 30%.

11. The World of Algorithms (Part II, Chapter 11, printed pages 37 to 50)

Q1 (2 marks, very short answer). Use Āryabhaṭa's division algorithm, gcd(m, n) = gcd(n, m mod n) (pages 48 to 49), to find gcd(252, 198). Show each step.

Answer: 18.

• 252 mod 198 = 54, so gcd(252, 198) = gcd(198, 54).
• 198 mod 54 = 36, so = gcd(54, 36). 54 mod 36 = 18, so = gcd(36, 18). 36 mod 18 = 0, so = gcd(18, 0) = 18.

Q2 (1 mark, multiple choice). Euclid's subtraction algorithm (page 46) is used on 91 and 35. What does it end with? (A) 5 (B) 7 (C) 13 (D) 35

Answer: (B). Keep subtracting the smaller number from the larger: (91, 35) → (56, 35) → (21, 35) → (21, 14) → (7, 14) → (7, 7), so gcd(91, 35) = 7.

12. Quadrilaterals (Part II, Chapter 12, printed pages 51 to 84)

Q1 (2 marks, very short answer). In parallelogram PQRS, ∠P = (5x − 2)° and ∠Q = (2x + 7)°. Find x and all four angles.

Answer: x = 25; ∠P = ∠R = 123°, ∠Q = ∠S = 57°.

• P and Q are adjacent angles of a parallelogram, so they add up to 180° (page 59): 7x + 5 = 180, x = 25.
• ∠P = 123° and ∠Q = 57°. Opposite angles are equal (page 57), so ∠R = 123° and ∠S = 57°.

Q2 (2 marks, very short answer). The diagonals of parallelogram ABCD meet at O. OA = (3y − 4) cm and OC = (y + 6) cm. Find y and the length of diagonal AC.

Answer: y = 5; AC = 22 cm. The diagonals of a parallelogram bisect each other (page 56), so OA = OC: 3y − 4 = y + 6, giving 2y = 10 and y = 5. Then OA = OC = 11 cm, and AC = 22 cm.

13. Two Variables, One Line (Part II, Chapter 13, printed pages 85 to 122)

Q1 (2 marks, very short answer). (a) Does the point (11, 10) lie on the line 3x − 2y = 13? (b) The point (k, −5) lies on the same line. Find k.

Answer: (a) yes; (b) k = 1.

• (a) 3(11) − 2(10) = 33 − 20 = 13 ✓, so (11, 10) is a solution and lies on the line.
• (b) 3k − 2(−5) = 13 gives 3k + 10 = 13, so k = 1.

Q2 (3 marks, short answer). A school spends exactly ₹720 on notebooks at ₹45 each and pens at ₹18 each. (a) Write a linear equation for x notebooks and y pens. (b) Find where its line meets each axis, and say what each point means. (c) If it buys 10 notebooks, how many pens does it buy?

Answer: (a) 45x + 18y = 720; (b) (16, 0) and (0, 40); (c) 15 pens.

• (b) y = 0 gives x = 16: 16 notebooks and no pens. x = 0 gives y = 40: 40 pens and no notebooks (intercepts, page 120).
• (c) 45(10) + 18y = 720, so 18y = 270 and y = 15.

14. Math of Space: Surface Area and Volume (Part II, Chapter 14, printed pages 123 to 148)

Q1 (2 marks, very short answer). A closed tin is a cylinder of radius 3.5 cm and height 12 cm. Taking π = 22/7, find its curved surface area and its volume.

Answer: 264 cm² and 462 cm³.

• Curved surface area = 2πrh = 2 × (22/7) × 3.5 × 12 = 264 cm² (page 128).
• Volume = πr²h = (22/7) × 3.5 × 3.5 × 12 = 462 cm³.

Q2 (5 marks, long answer). A wooden toy is a cone standing on a hemisphere of the same radius, 9 cm. The height of the cone is 12 cm. (a) Find the slant height of the cone. (b) Find the total outer surface area of the toy. (c) Find its volume. Leave your answers in terms of π.

Answer: (a) 15 cm; (b) 297π cm²; (c) 810π cm³.

• (a) l = √(9² + 12²) = √225 = 15 cm (slant height, page 133).
• (b) The flat faces are joined, so only the curved surfaces are outside. Cone: πrl = π × 9 × 15 = 135π. Hemisphere: 2πr² = 2π × 81 = 162π. Total = 297π cm².
• (c) Cone: (1/3)πr²h = (1/3)π × 81 × 12 = 324π. Hemisphere: (2/3)πr³ = (2/3)π × 729 = 486π. Total = 810π cm³ (volumes of a cone and a sphere, pages 132 to 138).

How to use these questions

• After each chapter: do its two questions with the book closed, then check your working against the answer, step by step, not just the final number.
• Before the half-yearly or annual paper: do all 28 in two sittings of about 90 minutes each, and mark the ones you got wrong. Go back to that chapter's guide for more practice.
• Show the working. In a 3 or 5-mark question most of the marks are for the steps: the identity or theorem you used, the substitution, and the unit at the end.

More practice, chapter by chapter, with eight questions each: Coordinates, Linear Polynomials, The World of Numbers, Algebraic Identities, Circles, Perimeter and Area, Probability, Sequences, Propositions, Understanding Data, Algorithms, Quadrilaterals, Two Variables, One Line, Surface Area and Volume. If you learnt the old chapter names, this page maps them to the new books.

Where CMDLearn fits

CMDLearn is building an AI tutor for Classes 1 to 10, in Hindi, Marathi and English, that explains one step, checks the child's working and leaves the writing to the child. It is still being built, and no child uses it yet. Join the waitlist to hear when there is something to try.

Sources

All checked on 4 November 2026.

• NCERT, Ganita Manjari, Textbook of Mathematics for Grade 9, Part I, Chapters 1 to 8: ncert.nic.in/textbook/pdf/iemh101.pdf to iemh108.pdf, and the contents page: iemh1ps.pdf
• NCERT, Ganita Manjari, Part II, Chapters 9 to 14: ncert.nic.in/textbook/pdf/iemh201.pdf to iemh206.pdf, and the contents page: iemh2ps.pdf
• CBSE, Mathematics, Class IX 2026-27: course structure (page 4), question paper design and internal assessment (page 13): cbseacademic.nic.in, Maths_SecP1IX_2026-27.pdf

Definitions, theorems and page numbers are from the documents above. Every question and answer on this page is our own, and every number is checked by a script.

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