CBSE Class 9 Maths Sample Paper 2026 (with Solutions) — Free PDF
This is a full-length CBSE Class 9 Mathematics paper built to the board's own pattern: 38 questions across five sections for 80 marks in 3 hours, covering
TL;DR: This is a full-length CBSE Class 9 Mathematics paper built to the board's own pattern: 38 questions across five sections for 80 marks in 3 hours, cove…
Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated
This is a full-length CBSE Class 9 Mathematics paper built to the board's own pattern: 38 questions across five sections for 80 marks in 3 hours, covering number systems, polynomials, coordinate geometry, linear equations, geometry, mensuration, statistics and probability. Section A is objective and Section E is case-based. Every question is followed by a fully worked solution showing each step, so you can check your method as well as your answer.
Board: CBSE · Class: Class 9 · Subject: Mathematics · Total: 80 marks · Time: 3 hours
Section A — Objective (1 mark each) — 20 marks
Questions 1–18 are multiple choice. Questions 19 and 20 are Assertion–Reason: mark (a) if both A and R are true and R explains A, (b) if both are true but R does not explain A, (c) if A is true and R is false, (d) if A is false and R is true.
Q1. Every rational number is (1 mark)
Answer: A real number. Rational numbers form a subset of the reals, but not every real number is rational — √2, for instance, is not.
Q2. The decimal expansion of an irrational number is (1 mark)
Answer: Non-terminating and non-recurring. A terminating or recurring expansion always represents a rational number.
Q3. The value of (√5 + √2)(√5 − √2) is (1 mark)
Answer: 3. Using (a + b)(a − b) = a² − b², the product is 5 − 2 = 3.
Q4. The degree of the polynomial 4x³ − 7x² + 2x − 9 is (1 mark)
Answer: 3 — the highest power of the variable appearing in the polynomial.
Q5. If p(x) = x² − 3x + 2, then p(1) equals (1 mark)
Answer: 0. p(1) = 1 − 3 + 2 = 0, so x = 1 is a zero of the polynomial.
Q6. The point (−3, 5) lies in which quadrant? (1 mark)
Answer: The second quadrant, where x is negative and y is positive.
Q7. The point (0, −4) lies on (1 mark)
Answer: The y-axis, 4 units below the origin. Any point with x = 0 lies on the y-axis.
Q8. The equation of a line parallel to the x-axis at a distance 3 units above it is (1 mark)
Answer: y = 3. Every point on such a line has the same y-coordinate.
Q9. In the linear equation 2x + 3y = 12, if x = 3 then y equals (1 mark)
Answer: 2. Substituting, 6 + 3y = 12, so 3y = 6 and y = 2.
Q10. Two angles of a triangle measure 45° and 65°. The third angle is (1 mark)
Answer: 70°. The angles of a triangle sum to 180°, so the third is 180 − 110 = 70°.
Q11. In a parallelogram, the diagonals (1 mark)
Answer: Bisect each other. They are not necessarily equal or perpendicular — those hold for rectangles and rhombuses respectively.
Q12. The sum of the interior angles of a quadrilateral is (1 mark)
Answer: 360°. A quadrilateral splits into two triangles, each contributing 180°.
Q13. Two triangles are congruent by the SSS rule when (1 mark)
Answer: All three sides of one triangle are equal to the corresponding three sides of the other.
Q14. The area of a triangle with base 10 cm and height 6 cm is (1 mark)
Answer: 30 cm². Area = ½ × base × height = ½ × 10 × 6 = 30 cm².
Q15. The volume of a cube of edge 4 cm is (1 mark)
Answer: 64 cm³. Volume = a³ = 4³ = 64 cm³.
Q16. The curved surface area of a cylinder of radius 7 cm and height 10 cm is (use π = 22/7) (1 mark)
Answer: 440 cm². CSA = 2πrh = 2 × (22/7) × 7 × 10 = 440 cm².
Q17. The mean of the first five natural numbers is (1 mark)
Answer: 3. Their sum is 1 + 2 + 3 + 4 + 5 = 15, and 15 ÷ 5 = 3.
Q18. A coin is tossed once. The probability of getting a tail is (1 mark)
Answer: 1/2. There are two equally likely outcomes and one of them is a tail.
Q19. Assertion (A): √2 is an irrational number. Reason (R): The square root of every prime number is irrational. (1 mark)
Answer: (a) Both true and R explains A. Since 2 is prime, the general result in R covers the particular case in A.
Q20. Assertion (A): The point (3, 0) lies on the x-axis. Reason (R): A point lies on the x-axis when its y-coordinate is zero. (1 mark)
Answer: (a) Both true and R explains A. The point has y = 0, which is exactly the condition stated in the reason.
Section B — Very Short Answer (2 marks each) — 10 marks
Answer all 5 questions. Each carries 2 marks.
Q1. Rationalise the denominator of 1/(√7 − √3). (2 marks)
Answer: Multiply numerator and denominator by the conjugate √7 + √3: 1/(√7 − √3) × (√7 + √3)/(√7 + √3) = (√7 + √3)/(7 − 3) = (√7 + √3)/4.
Q2. Find the value of k if x = 2 is a zero of the polynomial p(x) = x² + kx + 6. (2 marks)
Answer: If x = 2 is a zero then p(2) = 0, so 4 + 2k + 6 = 0. Hence 2k = −10 and k = −5.
Q3. Write two solutions of the linear equation 3x + 4y = 12. (2 marks)
Answer: Put x = 0: 4y = 12, so y = 3, giving (0, 3). Put y = 0: 3x = 12, so x = 4, giving (4, 0). Both pairs satisfy the equation, and there are infinitely many others.
Q4. In triangle ABC, AB = AC and angle B = 50°. Find angle A. (2 marks)
Answer: Since AB = AC the triangle is isosceles and the angles opposite the equal sides are equal, so angle C = angle B = 50°. Then angle A = 180 − 50 − 50 = 80°.
Q5. Find the total surface area of a cube of edge 5 cm. (2 marks)
Answer: A cube has six square faces, so TSA = 6a² = 6 × 25 = 150 cm².
Section C — Short Answer (3 marks each) — 18 marks
Answer all 6 questions. Each carries 3 marks.
Q1. Represent √3 on the number line. (3 marks)
Answer: Draw a number line and mark OA = 1 unit from the origin O. At A erect a perpendicular AB of 1 unit; then OB = √(1² + 1²) = √2 by the Pythagoras theorem. Now at B erect a perpendicular BC of 1 unit; then OC = √((√2)² + 1²) = √3. With O as centre and radius OC, draw an arc cutting the number line at P. The point P represents √3.
Q2. Factorise x³ − 3x² − 9x − 5. (3 marks)
Answer: Try x = −1: (−1) − 3(1) − 9(−1) − 5 = −1 − 3 + 9 − 5 = 0, so (x + 1) is a factor. Dividing gives x³ − 3x² − 9x − 5 = (x + 1)(x² − 4x − 5). The quadratic factors as (x − 5)(x + 1). Hence the full factorisation is (x + 1)²(x − 5).
Q3. If x + y = 12 and xy = 27, find the value of x³ + y³. (3 marks)
Answer: Use the identity x³ + y³ = (x + y)³ − 3xy(x + y). Substituting, x³ + y³ = 12³ − 3(27)(12) = 1728 − 972 = 756.
Q4. Prove that the angles opposite to equal sides of a triangle are equal. (3 marks)
Answer: Let triangle ABC have AB = AC, and draw AD as the bisector of angle A, meeting BC at D. In triangles ABD and ACD: AB = AC (given), angle BAD = angle CAD (by construction) and AD = AD (common). Hence the triangles are congruent by the SAS rule, and by CPCT angle B = angle C. So the angles opposite the equal sides are equal.
Q5. The sides of a triangle are 13 cm, 14 cm and 15 cm. Find its area using Heron's formula. (3 marks)
Answer: The semi-perimeter is s = (13 + 14 + 15)/2 = 21 cm. By Heron's formula, area = √(s(s−a)(s−b)(s−c)) = √(21 × 8 × 7 × 6) = √7056 = 84 cm².
Q6. The following are the marks of 10 students: 25, 30, 35, 30, 40, 45, 30, 50, 35, 40. Find the mean, median and mode. (3 marks)
Answer: The sum is 25 + 30 + 35 + 30 + 40 + 45 + 30 + 50 + 35 + 40 = 360, so the mean is 360/10 = 36. Arranged in order: 25, 30, 30, 30, 35, 35, 40, 40, 45, 50; with ten values the median is the mean of the 5th and 6th, that is (35 + 35)/2 = 35. The value 30 occurs three times, more than any other, so the mode is 30.
Section D — Long Answer (5 marks each) — 20 marks
Answer all 4 questions. Each carries 5 marks.
Q1. Prove that the diagonal of a parallelogram divides it into two congruent triangles, and hence that opposite sides of a parallelogram are equal. (5 marks)
Answer: Let ABCD be a parallelogram and join the diagonal AC. Since AB ∥ DC and AC is a transversal, angle BAC = angle DCA (alternate angles). Since AD ∥ BC and AC is a transversal, angle DAC = angle BCA (alternate angles). Also AC = CA is common. Hence triangles ABC and CDA are congruent by the ASA rule. By CPCT, AB = CD and BC = AD, so the opposite sides of a parallelogram are equal — which is what was to be proved.
Q2. A cone of height 24 cm has a curved surface area of 550 cm². Find its radius, slant height and volume. (Use π = 22/7) (5 marks)
Answer: Let the radius be r and the slant height l, with l = √(r² + h²) = √(r² + 576). CSA = πrl = 550, so (22/7)r√(r² + 576) = 550, giving r√(r² + 576) = 175. Squaring, r²(r² + 576) = 30625. Let u = r²: u² + 576u − 30625 = 0, so u = (−576 + √(331776 + 122500))/2 = (−576 + √454276)/2 = (−576 + 674)/2 = 49. Hence r = 7 cm, and l = √(49 + 576) = √625 = 25 cm. The volume is (1/3)πr²h = (1/3)(22/7)(49)(24) = 1232 cm³.
Q3. The angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all four angles, and state what kind of quadrilateral it cannot be. (5 marks)
Answer: Let the angles be 3x, 5x, 9x and 13x. Their sum is 360°, so 30x = 360 and x = 12. The angles are therefore 36°, 60°, 108° and 156°. Since no two angles are equal, and a parallelogram requires opposite angles to be equal, the quadrilateral cannot be a parallelogram — nor, for the same reason, a rectangle, rhombus or square. It is an ordinary irregular quadrilateral, and having one angle greater than 90° more than once it is not cyclic either, since 36 + 108 ≠ 180 and 60 + 156 ≠ 180.
Q4. The heights (in cm) of 20 students are 150, 152, 155, 160, 158, 162, 151, 157, 159, 156, 154, 156, 161, 153, 158, 157, 163, 166, 168, 155. Group them into classes of width 5 starting at 150, using the convention that a class includes its lower limit but not its upper. State the modal class and the class with the fewest students. (5 marks)
Answer: Sorting the values: 150, 151, 152, 153, 154, 155, 155, 156, 156, 157, 157, 158, 158, 159, 160, 161, 162, 163, 166, 168. Class 150–155 takes 150, 151, 152, 153, 154 — frequency 5. Class 155–160 takes 155, 155, 156, 156, 157, 157, 158, 158, 159 — frequency 9. Class 160–165 takes 160, 161, 162, 163 — frequency 4. Class 165–170 takes 166, 168 — frequency 2. The frequencies 5, 9, 4 and 2 total 20, which checks against the count. The modal class is 155–160, since it has the highest frequency of 9, and the class with the fewest students is 165–170 with a frequency of 2. The distribution is therefore concentrated around the middle of the range and thins out towards the taller heights.
Section E — Case-based questions (4 marks each) — 12 marks
Answer all 3 case-based questions. Each has sub-parts totalling 4 marks.
Q1. CASE STUDY 1 — A school wants to build a rectangular garden of length 20 m and breadth 15 m, and lay a path of uniform width 2 m all around it on the outside. (i) Find the area of the garden. (ii) Find the dimensions of the outer rectangle. (iii) Find the area of the path. (4 marks)
Answer: (i) Area of the garden = 20 × 15 = 300 m². (ii) The path adds 2 m on each of the four sides, so the outer rectangle measures (20 + 4) by (15 + 4), that is 24 m by 19 m. (iii) Area of the outer rectangle = 24 × 19 = 456 m², so the area of the path = 456 − 300 = 156 m².
Q2. CASE STUDY 2 — A water tank in the shape of a cylinder has a radius of 3.5 m and a height of 4 m. (Use π = 22/7) (i) Find its volume in cubic metres. (ii) Find how many litres it holds. (iii) Find the cost of painting its curved surface at ₹20 per m². (4 marks)
Answer: (i) Volume = πr²h = (22/7)(3.5)²(4) = (22/7)(12.25)(4) = 154 m³. (ii) One cubic metre is 1000 litres, so the tank holds 154,000 litres. (iii) Curved surface area = 2πrh = 2 × (22/7) × 3.5 × 4 = 88 m², so the cost is 88 × 20 = ₹1,760.
Q3. CASE STUDY 3 — In a survey of 60 families, the number of children per family was recorded: 0 children in 5 families, 1 in 12, 2 in 25, 3 in 13 and 4 in 5. (i) Find the total number of children. (ii) Find the mean number of children per family. (iii) A family is chosen at random — find the probability that it has at least two children. (4 marks)
Answer: (i) Total children = 0(5) + 1(12) + 2(25) + 3(13) + 4(5) = 0 + 12 + 50 + 39 + 20 = 121. (ii) Mean = 121/60 ≈ 2.02 children per family. (iii) Families with at least two children number 25 + 13 + 5 = 43, so the probability is 43/60 ≈ 0.717.
FAQs
What are the main chapters in Class 9 Maths?
Number systems, polynomials, coordinate geometry, linear equations, triangles, circles, statistics, and probability.
Is it necessary to memorize formulas for the exam?
Yes, formulas for area, volume, and algebraic identities must be memorized for quick problem-solving.
How can I improve my geometry skills?
Practice construction problems regularly and understand the underlying theorems before solving numerical problems.
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