CBSE Class 9 · Mathematics (NCERT ‘Ganita Manjari’) · Max Marks 80 · Time 3 hrs · Exam-level hard
CBSE Class 9 Maths — Model Question Paper (Set 1) · Max Marks 80 · Time 3 hrs
NCERT ‘Ganita Manjari’ (2026-27) pattern · exam-level hard. Answers & marking scheme in the ✍️ Answers tab.
General Instructions: (i) Five sections A–E, 38 compulsory questions. (ii) A (Q1–20) 1 mark; B (Q21–25) 2 marks; C (Q26–31) 3 marks; D (Q32–35) 5 marks; E (Q36–38) 4 marks. (iii) Internal choices are given in some questions. (iv) Use π = 22/7 unless stated. Draw neat figures where needed.
SECTION A — Objective [20 × 1 = 20]
Q1. The point (−3, 5) lies in which quadrant?
- (i) Quadrant II
- (ii) Quadrant I
- (iii) Quadrant III
- (iv) Quadrant IV
Q2. The distance between the points (0, 0) and (3, 4) is:
- (i) 5
- (ii) 7
- (iii) 1
- (iv) √7
Q3. The degree of the polynomial 4x − 3 is:
Q4. The decimal expansion of 3/8 is:
- (i) terminating
- (ii) non-terminating repeating
- (iii) non-terminating non-repeating
- (iv) not defined
Q5. (a + b)² is equal to:
- (i) a² + 2ab + b²
- (ii) a² + b²
- (iii) a² − 2ab + b²
- (iv) 2ab
Q6. One factor of x² + 7x + 12 is:
- (i) (x + 3)
- (ii) (x − 3)
- (iii) (x + 5)
- (iv) (x − 4)
Q7. √2 is:
- (i) an irrational number
- (ii) a rational number
- (iii) an integer
- (iv) a natural number
Q8. The angle in a semicircle is:
- (i) 90°
- (ii) 180°
- (iii) 60°
- (iv) 45°
Q9. The sum of opposite angles of a cyclic quadrilateral is:
- (i) 180°
- (ii) 90°
- (iii) 360°
- (iv) 120°
Q10. The circumference of a circle of radius 7 cm (π = 22/7) is:
- (i) 44 cm
- (ii) 22 cm
- (iii) 14 cm
- (iv) 154 cm
Q11. The area of a triangle with sides 3, 4, 5 cm is:
- (i) 6 cm²
- (ii) 12 cm²
- (iii) 10 cm²
- (iv) 7.5 cm²
Q12. The area of a circle of radius r is:
- (i) πr²
- (ii) 2πr
- (iii) πd
- (iv) πr
Q13. The probability of getting a head on a fair coin is:
- (i) 1/2
- (ii) 1/4
- (iii) 1
- (iv) 0
Q14. The n-th term of an AP with first term a and common difference d is:
- (i) a + (n − 1)d
- (ii) a·rⁿ⁻¹
- (iii) n(n+1)/2
- (iv) a + nd
Q15. The common ratio of the GP 3, 6, 12, 24, … is:
Q16. The sum of the first n natural numbers is:
- (i) n(n + 1)/2
- (ii) n²
- (iii) 2n − 1
- (iv) n(n − 1)/2
Q17–20 (Assertion–Reason): (a) Both true, R explains A; (b) Both true, R not the explanation; (c) A true, R false; (d) A false, R true.
Q17. Assertion: (a + b)² = a² + b². Reason: (a + b)² = a² + 2ab + b².
Q18. Assertion: The midpoint of (2, 4) and (6, 8) is (4, 6). Reason: The midpoint of a segment is the average of the coordinates of its endpoints.
Q19. Assertion: √9 is an irrational number. Reason: An irrational number cannot be written as p/q with integers p, q (q ≠ 0).
Q20. Assertion: In an AP the difference between consecutive terms is constant. Reason: In a GP the ratio between consecutive terms is constant.
SECTION B — Very Short Answer [5 × 2 = 10]
Q21. Using a suitable identity, find the value of (102)². [2]
Q22. Find three rational numbers between 1/4 and 1/2. [2]
Q23. Find the 10th term of the AP 2, 5, 8, 11, … [2]
Q24. A chord of length 8 cm is at a distance of 3 cm from the centre of a circle. Find the radius. [2]
Q25. A die is rolled once. Find the probability of getting a prime number. [2]
SECTION C — Short Answer [6 × 3 = 18]
Q26. Expand (2x + 3y)² using an identity, and find its value when x = 1 and y = 1. [3]
Q27. Prove that √5 is an irrational number. [3]
Q28. Find the area of a triangle with sides 5 cm, 12 cm and 13 cm using Heron’s formula, and verify it is right-angled. [3]
Q29. Plot (conceptually) A(1, 2), B(4, 2) and C(4, 6). Find AB, BC and AC, show the triangle is right-angled, and find its area. [3]
Q30. The 4th term of an AP is 11 and the 7th term is 20. Find the first term, the common difference, and write the AP. [3]
Q31. Convert the repeating decimal 0.͖36 (0.363636…) into the form p/q. [3]
SECTION D — Long Answer [4 × 5 = 20]
Q32. (a) Expand (2x − y + 3z)² using the identity for (a + b + c)². (b) Factorise 8x³ + 27y³. [5]
OR — Factorise completely: (a) x² + (5/6)x + 1/6, and (b) p³ + q³ + 27r³ − 9pqr.
Q33. Prove that the angle subtended by an arc at the centre of a circle is double the angle subtended by it at any point on the remaining part of the circle. [5]
OR — Prove that equal chords of a circle are equidistant from the centre.
Q34. A circular park has radius 21 m. A path of width 3.5 m runs around it on the outside. Find the area of the path (π = 22/7). [5]
OR — Find the area of a sector of angle 60° and of the corresponding minor segment in a circle of radius 14 cm (π = 22/7, √3 = 1.73).
Q35. A person saves ₹200 in the first month and increases the saving by ₹50 each month. (a) Find the saving in the 12th month. (b) Find the total saved in one year. [5]
OR — Bacteria in a culture double every hour, starting from 30. Find the count after the 3rd, 5th and n-th hour, and state whether this is an AP or a GP with reason.
SECTION E — Case / Source-based [3 × 4 = 12]
Q36. On a graph, a rectangular garden has corners O(0, 0), A(8, 0), B(8, 6) and C(0, 6).
- (i) In which quadrant does the whole garden lie? [1]
- (ii) Find the length of the diagonal OB. [1]
- (iii) Find the midpoint of the diagonal AC. [1]
- (iv) Find the perimeter of the garden. [1]
Q37. A circular flower-bed has a chord PQ of length 24 cm drawn 5 cm from its centre O.
- (i) Find the radius of the circle. [1]
- (ii) What is the length of the perpendicular from O to PQ, and what does it do to PQ? [1]
- (iii) What is the longest possible chord of this circle? [1]
- (iv) If another chord is 12 cm from the centre, is it longer or shorter than PQ? Why? [1]
Q38. In a survey of 40 students, favourite sports were: Cricket 16, Football 10, Hockey 8, Tennis 6.
- (i) Find the probability that a randomly chosen student likes Cricket. [1]
- (ii) Find the probability that a student does NOT like Football. [1]
- (iii) If the school has 800 students, estimate how many like Hockey. [1]
- (iv) What is the sum of the probabilities of all four sports? [1]
Model Answers & Marking Scheme (Set 1)
Value-points; correct method in the student’s own words earns full marks.
Section A (1 mark each)
1 — (i) Quadrant II | 2 — (i) 5 | 3 — (i) 1 | 4 — (i) terminating | 5 — (i) a²+2ab+b² | 6 — (i) (x+3) | 7 — (i) irrational | 8 — (i) 90° | 9 — (i) 180° | 10 — (i) 44 cm | 11 — (i) 6 cm² | 12 — (i) πr² | 13 — (i) 1/2 | 14 — (i) a+(n−1)d | 15 — (i) 2 | 16 — (i) n(n+1)/2
17 — (d) A false, R true. 18 — (a) both true, R explains A. 19 — (d) A false (√9 = 3 is rational), R true. 20 — (b) both true, R not the explanation.
Section B (2 marks each)
21. (102)² = (100 + 2)² = 100² + 2·100·2 + 2² = 10000 + 400 + 4 = 10404.
22. Write 1/4 = 6/24 and 1/2 = 12/24. Three rationals between them: 7/24, 8/24 (= 1/3), 9/24 (= 3/8).
23. a = 2, d = 3, t₁₀ = a + 9d = 2 + 27 = 29.
24. Half-chord = 4 cm, distance = 3 cm. Radius = √(4² + 3²) = √25 = 5 cm.
25. Primes on a die are {2, 3, 5} = 3 outcomes. P(prime) = 3/6 = 1/2.
Section C (3 marks each)
26. (2x + 3y)² = 4x² + 12xy + 9y². At x = 1, y = 1: 4 + 12 + 9 = 25 (= (2+3)² ✓).
27. Assume √5 = p/q in lowest terms ⇒ 5q² = p² ⇒ p² (hence p) is a multiple of 5; put p = 5k ⇒ 5q² = 25k² ⇒ q² = 5k² ⇒ q is a multiple of 5. Then p, q share factor 5 — contradiction. So √5 is irrational.
28. s = (5+12+13)/2 = 15; area = √[15(15−5)(15−12)(15−13)] = √(15·10·3·2) = √900 = 30 cm². Since 5² + 12² = 13², it is right-angled (and ½·5·12 = 30 cm² ✓).
29. AB = |4 − 1| = 3 (horizontal); BC = |6 − 2| = 4 (vertical); AC = √(3² + 4²) = 5. Since AB ⊥ BC, the right angle is at B. Area = ½ × 3 × 4 = 6 sq units.
30. a + 3d = 11 and a + 6d = 20 ⇒ 3d = 9 ⇒ d = 3, a = 2. AP: 2, 5, 8, 11, 14, …
31. x = 0.3636… ⇒ 100x = 36.3636… ⇒ 99x = 36 ⇒ x = 36/99 = 4/11.
Section D (5 marks each)
32. (a) (2x − y + 3z)² = 4x² + y² + 9z² − 4xy − 6yz + 12zx. (b) 8x³ + 27y³ = (2x)³ + (3y)³ = (2x + 3y)(4x² − 6xy + 9y²). OR (a) x² + (5/6)x + 1/6 = (x + 1/2)(x + 1/3); (b) p³ + q³ + 27r³ − 9pqr = (p + q + 3r)(p² + q² + 9r² − pq − 3qr − 3rp).
33. Standard proof: join the point to the centre and extend; using the exterior-angle property of the two isosceles triangles (two sides being radii), the central angle = 2 × the inscribed angle. OR — equal chords AB = CD give ΔOAB ≅ ΔOCD (SSS), so the perpendiculars from O (the altitudes) are equal, i.e. the chords are equidistant from O.
34. Outer radius = 21 + 3.5 = 24.5 m. Path area = π(24.5² − 21²) = (22/7)(24.5 − 21)(24.5 + 21) = (22/7)(3.5)(45.5) = 500.5 m². OR — sector (60°, r = 14): area = (22/7)(14²)(60/360) = 102.67 cm²; triangle (equilateral, side 14) = (√3/4)(14²) = 84.77 cm²; minor segment = 102.67 − 84.77 ≈ 17.9 cm².
35. AP with a = 200, d = 50. (a) 12th month = 200 + 11×50 = ₹750. (b) Total = (12/2)(200 + 750) = 6 × 950 = ₹5700. OR — GP with a = 30, r = 2: after 3rd hour 30×2³ = 240; after 5th hour 30×2⁵ = 960; after n-th hour 30·2ⁿ. It is a GP (constant ratio 2), not an AP.
Section E (4 marks each)
36. (i) Quadrant I (all coordinates positive). (ii) OB = √(8² + 6²) = √100 = 10 units. (iii) Midpoint of AC = ((8+0)/2, (0+6)/2) = (4, 3). (iv) Perimeter = 2(8 + 6) = 28 units.
37. (i) Radius = √(12² + 5²) = √169 = 13 cm (half-chord 12, distance 5). (ii) Perpendicular from O = 5 cm; it bisects PQ. (iii) Longest chord = diameter = 26 cm. (iv) A chord 12 cm from the centre is shorter than PQ — the farther a chord is from the centre, the shorter it is.
38. (i) P(Cricket) = 16/40 = 2/5 (0.4). (ii) P(not Football) = (40 − 10)/40 = 30/40 = 3/4. (iii) Hockey estimate = (8/40) × 800 = 160 students. (iv) Sum of all probabilities = (16+10+8+6)/40 = 40/40 = 1.