Class 10 Maths Half-Yearly Model Paper 2026-27 (80 Marks) with Solutions, English Medium | RBSE

Class 10 Mathematics (Code 09) — Half-Yearly Model Paper 2026-27 (English Medium)

NCERTClasses Team

General Instructions

1. All questions are compulsory. The marks for each question are given in brackets against the question number.
2. Section A has 20 objective questions (1 mark each), Section B has 8 questions (2 marks each), Section C has 6 questions (3 marks each), Section D has 4 questions (4 marks each) and Section E has 2 questions (5 marks each).
3. Sections D and E have internal choice (OR); attempt only one of the two alternatives.
4. The unit name shown in brackets with each question is given for practice purposes only.
5. Show all calculations clearly; the use of a calculator is not permitted.

Section A (Objective Questions) — Questions 1 to 20, 1 mark each

1. If the HCF of two numbers is 12 and their product is 1800, then their LCM is: [Real Numbers] [1]
(A) 120(B) 150(C) 180(D) 21600

2. If α, β are the zeroes of the polynomial 2x² − 7x + 3, then the value of α + β + αβ is: [Polynomials] [1]
(A) 2(B) 4(C) 5(D) 7/2

3. The pair of equations 2x + 3y = 7 and 4x + 6y = k has infinitely many solutions if the value of k is: [Pair of Linear Equations in Two Variables] [1]
(A) 7(B) 14(C) 21(D) 28

4. If the roots of the equation x² − 3x + k = 0 are real and equal, then k = [Quadratic Equations] [1]
(A) 4/9(B) 3/2(C) 9/4(D) 9

5. The 15th term of the A.P. 7, 10, 13, … is: [Arithmetic Progressions] [1]
(A) 45(B) 46(C) 49(D) 52

6. If △ABC ∼ △DEF and AB : DE = 3 : 5, then the ratio of areas ar(△ABC) : ar(△DEF) is: [Triangles] [1]
(A) 3 : 5(B) 5 : 3(C) 9 : 25(D) 25 : 9

7. The radius of a circle is 5 cm. The length of the tangent drawn to the circle from a point 13 cm away from the centre is: [Circles] [1]
(A) 8 cm(B) 10 cm(C) 12 cm(D) 13 cm

8. The distance between the points (2, 3) and (−1, 7) is: [Coordinate Geometry] [1]
(A) 3(B) 4(C) 5(D) 7

9. If 5 tan θ = 4, then the value of (5 sin θ − 3 cos θ)/(5 sin θ + 2 cos θ) is: [Introduction to Trigonometry] [1]
(A) 1/6(B) 1/5(C) 5/6(D) 6

10. The area of a sector of angle 90° of a circle of radius 7 cm (π = 22/7) is: [Areas Related to Circles] [1]
(A) 19.25 cm²(B) 38.5 cm²(C) 77 cm²(D) 154 cm²

11. The volume of a cone of radius 3.5 cm and height 12 cm (π = 22/7) is: [Surface Areas and Volumes] [1]
(A) 77 cm³(B) 154 cm³(C) 308 cm³(D) 462 cm³

12. The formula l + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] × h is used to calculate which of the following? [Statistics] [1]
(A) Mean(B) Median(C) Mode(D) Range

13. When a die is thrown once, the probability of getting a prime number is: [Probability] [1]
(A) 1/6(B) 1/3(C) 1/2(D) 2/3

14. A cubic polynomial can have at most ______ zeroes. [Polynomials] [1]

15. The ratio of the sides of two similar triangles is 4 : 9. The ratio of their corresponding altitudes (heights) is ______. [Triangles] [1]

16. The tangent at any point of a circle is ______ to the radius through the point of contact. [Circles] [1]

17. The mid-point of the line segment joining the points (−2, 6) and (8, −2) is ______. [Coordinate Geometry] [1]

18. If the probability of an event E is P(E) = 0.35, then P(not E) = ______. [Probability] [1]

19. Assertion-Reason question: [Arithmetic Progressions] [1]
Assertion (A): The 11th term of the A.P. −3, −1/2, 2, … is 22.
Reason (R): The nth term of an A.P. is aₙ = a + (n − 1)d.
(A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.

20. Case-based question: Rohan stands 15 m away from the foot of a vertical flagpole. He observes the top of the pole at an angle of elevation of 30°. (Neglect the height of his eyes.) The height of the pole is: [Some Applications of Trigonometry] [1]
(A) 5√3 m(B) 15√3 m(C) 15 m(D) 7.5 m

Section B — Questions 21 to 28, 2 marks each

21. Find the zeroes of the polynomial 6x² − 7x − 3 and verify the relationship between the zeroes and the coefficients. [Polynomials] [2]

22. In △ABC, D lies on side AB and E lies on side AC such that DE ∥ BC. If AD = 3 cm, DB = 5 cm and AE = 4.5 cm, find EC. [Triangles] [2]

23. Find the value of y for which the distance between the points P(2, −3) and Q(10, y) is 10 units. [Coordinate Geometry] [2]

24. Evaluate: 2 tan² 45° + cos² 30° − sin² 60° [Introduction to Trigonometry] [2]

25. If sin(A + B) = 1 and cos(A − B) = √3/2, 0° < A + B ≤ 90°, A > B, find A and B. [Introduction to Trigonometry] [2]

26. Write the modal class and the median class of the following frequency distribution: [Statistics] [2]

27. The mean of a distribution is 22 and its median is 21. Find its mode. [Statistics] [2]

28. A bag contains 3 red, 5 black and 2 white balls. One ball is drawn at random from the bag. Find the probability that the ball is (i) black, (ii) not white. [Probability] [2]

Section C — Questions 29 to 34, 3 marks each

29. Find the HCF and LCM of 120 and 144 by the prime factorisation method, and verify that HCF × LCM = product of the two numbers. [Real Numbers] [3]

30. The product of two consecutive positive integers is 306. Find the integers. [Quadratic Equations] [3]

31. How many terms of the A.P. 24, 21, 18, … must be taken so that their sum is 78? Explain the double answer. [Arithmetic Progressions] [3]

32. Find the coordinates of the point which divides the line segment joining A(−1, 7) and B(4, −3) internally in the ratio 2 : 3. [Coordinate Geometry] [3]

33. Prove that: (cosec θ − cot θ)² = (1 − cos θ)/(1 + cos θ) [Introduction to Trigonometry] [3]

34. Find the mode of the following distribution: [Statistics] [3]

Section D — Questions 35 to 38, 4 marks each (with internal choice)

35. If 2 is added to both the numerator and the denominator of a fraction, it becomes 9/11, and if 3 is added to both, it becomes 5/6. Find the fraction. [Pair of Linear Equations in Two Variables] [4]

OR

A boat takes 10 hours to go 30 km upstream and 44 km downstream. If it goes 40 km upstream and 55 km downstream, it takes 13 hours. Find the speed of the stream and the speed of the boat in still water.

36. Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact. [Circles] [4]

OR

From an external point P, two tangents PA and PB are drawn to a circle with centre O (A and B are the points of contact). Prove that ∠APB + ∠AOB = 180°.

37. From the top of a 60 m high building, the angles of depression of the top and the foot of a vertical tower are 30° and 60° respectively. Find the height of the tower. (The building and the tower stand on level ground.) [Some Applications of Trigonometry] [4]

OR

The angles of elevation of the top of a tower from two points on the same straight line through its foot, on the same side of the tower, at distances of 4 m and 9 m from the foot, are complementary. Prove that the height of the tower is 6 m.

38. A chord of a circle of radius 12 cm subtends an angle of 60° at the centre. Find the area of the corresponding minor segment. (Use π = 3.14 and √3 = 1.73) [Areas Related to Circles] [4]

OR

In a square of side 14 cm, four quadrants (quarter circles) of radius 7 cm are drawn inside the square, each with a vertex of the square as centre. Find the area of the part of the square not covered by the quadrants. (π = 22/7)

Section E — Questions 39 and 40, 5 marks each (with internal choice)

39. Find the mean of the following distribution by the step-deviation method: [Statistics] [5]

OR

Find the median of the following distribution: (same table)

40. A solid metallic sphere of radius 6 cm is melted and made into cones, each of radius 3 cm and height 4 cm. (i) How many cones are formed? (ii) Find the curved surface area of one cone in terms of π. [Surface Areas and Volumes] [5]

OR

A toy is in the shape of a cone of radius 3.5 cm mounted on a hemisphere of the same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy. (π = 22/7)

Unit-wise Marks Table

Answer Key and Marking Scheme

Section A (1 mark each)

1. (B) 150 — LCM = 1800 ÷ 12 = 150
2. (C) 5 — α+β = 7/2, αβ = 3/2, sum = 5
3. (B) 14 — 2/4 = 3/6 = 7/k ⇒ k = 14
4. (C) 9/4 — D = 9 − 4k = 0
5. (C) 49 — 7 + 14×3 = 49
6. (C) 9 : 25 — ratio of areas = square of ratio of sides
7. (C) 12 cm — √(13² − 5²) = 12
8. (C) 5 — √(9 + 16) = 5
9. (A) 1/6 — dividing numerator and denominator by cos θ gives (4 − 3)/(4 + 2)
10. (B) 38.5 cm² — (90/360)×(22/7)×49
11. (B) 154 cm³ — (1/3)×(22/7)×(3.5)²×12
12. (C) Mode
13. (C) 1/2 — favourable outcomes 2, 3, 5 ⇒ 3/6
14. 3
15. 4 : 9
16. perpendicular (at 90°)
17. (3, 2)
18. 0.65
19. (A) — a = −3, d = 5/2, a₁₁ = −3 + 10×5/2 = 22; R is the correct formula
20. (A) 5√3 m — h = 15 tan 30° = 15/√3 = 5√3

Section B (2 marks each)

21. 6x² − 7x − 3 = (3x + 1)(2x − 3) ⇒ zeroes −1/3 and 3/2 (1 mark). Sum = −1/3 + 3/2 = 7/6 = −(−7)/6 ✓; product = −1/2 = −3/6 ✓ (1 mark).

22. By BPT, AD/DB = AE/EC (1 mark) ⇒ 3/5 = 4.5/EC ⇒ EC = 7.5 cm (1 mark).

23. (10 − 2)² + (y + 3)² = 10² (1 mark) ⇒ (y + 3)² = 36 ⇒ y = 3 or y = −9 (1 mark).

24. 2(1) + (√3/2)² − (√3/2)² = 2 + 3/4 − 3/4 = 2 (values 1 + simplification 1).

25. sin(A+B) = 1 ⇒ A + B = 90°; cos(A−B) = √3/2 ⇒ A − B = 30° (1 mark) ⇒ A = 60°, B = 30° (1 mark).

26. Highest frequency is 12 ⇒ modal class 20–30 (1 mark). N = 35, N/2 = 17.5; cumulative frequencies 4, 11, 23 ⇒ median class 20–30 (1 mark).

27. Mode = 3 × median − 2 × mean = 63 − 44 = 19 (formula 1 + value 1).

28. Total balls = 10. (i) P(black) = 5/10 = 1/2 (1 mark) (ii) P(not white) = 8/10 = 4/5 (1 mark).

Section C (3 marks each)

29. 120 = 2³ × 3 × 5; 144 = 2⁴ × 3² (1 mark). HCF = 2³ × 3 = 24; LCM = 2⁴ × 3² × 5 = 720 (1 mark). 24 × 720 = 17280 = 120 × 144 ✓ (1 mark).

30. Integers x, x + 1: x(x + 1) = 306 ⇒ x² + x − 306 = 0 (1 mark) ⇒ (x + 18)(x − 17) = 0 (1 mark) ⇒ x = 17 (positive). The integers are 17 and 18 (1 mark).

31. a = 24, d = −3; Sₙ = n/2[48 − 3(n − 1)] = 78 ⇒ n² − 17n + 52 = 0 (1 mark) ⇒ (n − 4)(n − 13) = 0 ⇒ n = 4 or 13 (1 mark). Explanation: the sum of the first 4 terms (24, 21, 18, 15) is 78; the sum of the 5th to 13th terms (12, 9, 6, 3, 0, −3, −6, −9, −12) is zero, so the sum of 13 terms is also 78 (1 mark).

32. x = (2×4 + 3×(−1))/5 = 1 (1.5 marks); y = (2×(−3) + 3×7)/5 = 3 (1.5 marks). The point is (1, 3).

33. LHS = (1/sin θ − cos θ/sin θ)² = (1 − cos θ)²/sin²θ (1 mark) = (1 − cos θ)²/[(1 − cos θ)(1 + cos θ)] (1 mark) = (1 − cos θ)/(1 + cos θ) = RHS (1 mark).

34. N = 45; modal class 30–40 (highest frequency 15), l = 30, f₁ = 15, f₀ = 9, f₂ = 6, h = 10 (1 mark). Mode = 30 + (15 − 9)/(30 − 9 − 6) × 10 (1 mark) = 30 + 4 = 34 (1 mark).

Section D (4 marks each)

35 (i). Let the fraction be x/y: (x + 2)/(y + 2) = 9/11 ⇒ 11x − 9y = −4; (x + 3)/(y + 3) = 5/6 ⇒ 6x − 5y = −3 (2 marks). Solving, x = 7, y = 9 (1 mark). Fraction = 7/9 (check: 9/11, 10/12 ✓) (1 mark).
OR Let the speed in still water be x and the speed of the stream be y: 30/(x−y) + 44/(x+y) = 10; 40/(x−y) + 55/(x+y) = 13. Put u = 1/(x−y), v = 1/(x+y): 30u + 44v = 10, 40u + 55v = 13 (1.5 marks) ⇒ v = 1/11, u = 1/5 (1 mark) ⇒ x + y = 11, x − y = 5 ⇒ speed of boat in still water 8 km/h, speed of stream 3 km/h (1.5 marks).

36 (i). Given: circle with centre O, tangent XY, point of contact P. To prove: OP ⊥ XY (1 mark). Take any other point Q on XY; Q lies outside the circle, so OQ > OP (1 mark). This is true for every such Q, so OP is the shortest distance from O to the line (1 mark). The shortest distance from a point to a line is the perpendicular, so OP ⊥ XY (1 mark).
OR OA ⊥ PA, OB ⊥ PB ⇒ ∠OAP = ∠OBP = 90° (1 mark). The sum of the angles of quadrilateral OAPB is 360° (1 mark) ⇒ ∠APB + ∠AOB = 360° − 90° − 90° (1 mark) = 180° (1 mark).

37 (i). Building AB = 60 m, tower CD; draw DE perpendicular to AB from D. Angles of depression ⇒ ∠BCA = 60°, ∠BDE = 30° (1 mark). In △ABC, AC = 60/tan 60° = 20√3 m (1 mark). DE = AC = 20√3, BE = DE tan 30° = 20 m (1 mark). Height of tower CD = AE = 60 − 20 = 40 m (1 mark).
OR Let the height of the tower be h and the angles be θ and (90° − θ): tan θ = h/4 and tan(90° − θ) = cot θ = h/9 (2 marks). Multiplying, 1 = h²/36 (1 mark) ⇒ h² = 36 ⇒ h = 6 m (1 mark).

38 (i). Area of sector = (60/360) × 3.14 × 12² = 75.36 cm² (1.5 marks). Area of triangle = (√3/4) × 12² = 62.28 cm² (1.5 marks). Minor segment = 75.36 − 62.28 = 13.08 cm² (1 mark).
OR Area of square = 14² = 196 cm² (1 mark). Total area of the four quadrants = 4 × (1/4) × (22/7) × 7² = 154 cm² (2 marks). Remaining area = 196 − 154 = 42 cm² (1 mark).

Section E (5 marks each)

39 (i). Class marks x = 10, 30, 50, 70, 90, 110; take a = 50, h = 20, u = (x − 50)/20 = −2, −1, 0, 1, 2, 3 (1 mark). fu = −10, −8, 0, 12, 14, 9 ⇒ Σfu = 17, Σf = 50 (2 marks). Mean = a + h × Σfu/Σf = 50 + 20 × 17/50 (1 mark) = 56.8 (1 mark).
OR Cumulative frequencies 5, 13, 28, 40, 47, 50 (1 mark); N/2 = 25 ⇒ median class 40–60 (1 mark); l = 40, cf = 13, f = 15, h = 20 (1 mark). Median = 40 + (25 − 13)/15 × 20 (1 mark) = 56 (1 mark).

40 (i). Volume of sphere = (4/3)π(6)³ = 288π cm³ (1 mark). Volume of one cone = (1/3)π(3)²(4) = 12π cm³ (1 mark). Number of cones = 288π/12π = 24 (1 mark). Slant height l = √(3² + 4²) = 5 cm (1 mark); curved surface area = πrl = π × 3 × 5 = 15π cm² (1 mark).
OR Height of cone = 15.5 − 3.5 = 12 cm (1 mark); l = √(12² + 3.5²) = √156.25 = 12.5 cm (1 mark). TSA = πrl + 2πr² (1 mark) = (22/7)(3.5)(12.5) + 2(22/7)(3.5)² = 137.5 + 77 (1 mark) = 214.5 cm² (1 mark).