X Mat SP 1

CBSE CLASS X MATHEMATICS

MODEL QUESTION PAPER - 1

Class: X	Academic Session: 2026-27
Subject: Mathematics (Standard / Basic)	Time Allowed: 1.5 Hours
Coverage: Chapters 1 to 9	Max. Marks: 40

SECTION A: MULTIPLE CHOICE QUESTIONS [10 MARKS]

Q1. If two positive integers a and b are written as a = x³y² and b = xy³, where x, y are prime numbers, then HCF(a, b) is: [1]

(A) xy

(B) xy²

(C) x³y³

(D) x²y²

Q2. If one zero of the quadratic polynomial x² + 3x + k is 2, then the value of k is: [1]

(A) 10

(B) −10

(C) −7

(D) 7

Q3. The pair of equations x + 2y + 5 = 0 and −3x − 6y + 1 = 0 has: [1]

(A) a unique solution

(B) exactly two solutions

(C) infinitely many solutions

(D) no solution

Q4. The roots of the quadratic equation x² − 0.04 = 0 are: [1]

(A) ±0.2

(B) ±0.02

(C) 0.4

(D) 2

Q5. The 11^th term of the A.P. −3, −1/2, 2, ... is: [1]

(A) 28

(B) 22

(C) −38

(D) 25

Q6. In ΔABC, DE || BC such that AD = 3 cm, DB = 5 cm, and AC = 5.6 cm. The length of AE is: [1]

(A) 2.1 cm

(B) 3.1 cm

(C) 1.8 cm

(D) 2.4 cm

Q7. The distance of the point P(−6, 8) from the origin is: [1]

(A) 8 units

(B) 2√7 units

(C) 10 units

(D) 6 units

Q8. If sin θ + cos θ = √2 cos θ, then the value of tan θ is: [1]

(A) √2 − 1

(B) √2 + 1

(C) 1 / √2

(D) √3

Q9. If the angle of elevation of the top of a tower from a point on the ground 30 m away from its foot is 30°, then the height of the tower is: [1]

(A) 10√3 m

(B) 30√3 m

(C) 20 m

(D) 10 m

Q10. The midpoint of the line segment joining points A(3, 4) and B(k, 6) is P(1, 5). The value of k is: [1]

(A) 1

(B) −1

(C) 2

(D) −2

SECTION B: FILL IN THE BLANKS [10 MARKS]

Q11. The product of HCF and LCM of two numbers is equal to the product of the ____________________. [1]

Q12. A quadratic equation ax² + bx + c = 0 has two equal real roots if its discriminant D is equal to ____________________. [1]

Q13. The n^th term of an A.P. whose first term is a and common difference is d is given by a_n = ____________________. [1]

Q14. All ____________________ triangles are similar to each other. [1]

Q15. The coordinates of a point lying on the y-axis are of the form ____________________. [1]

Q16. Value of (sin² 30° + cos² 30°) is equal to ____________________. [1]

Q17. The ratio of the length of a vertical rod and its shadow is 1 : √3. The angle of elevation of the sun is ____________________. [1]

Q18. If the lines given by 3x + 2ky = 2 and 2x + 5y + 1 = 0 are parallel, then value of k is ____________________. [1]

Q19. The maximum number of zeroes a cubic polynomial can have is ____________________. [1]

Q20. Distance between points (a, b) and (−a, −b) is ____________________. [1]

SECTION C: SHORT ANSWER QUESTIONS [10 MARKS]

Q21. Prove that 5 − √3 is an irrational number, given that √3 is irrational. [2]

Q22. Find the zeroes of the quadratic polynomial 6x² − 3 − 7x and verify the relationship between the zeroes and coefficients. [2]

Q23. Which term of the A.P. 21, 18, 15, ... is −81? [2]

Q24. Find the ratio in which the y-axis divides the line segment joining the points (5, −6) and (−1, −4). [2]

Q25. Prove that: (sec A + tan A)(1 − sin A) = cos A. [2]

SECTION D: LONG ANSWER & CASE STUDY QUESTIONS [10 MARKS]

Q26. [5]
(a) The sum of the 4^th and 8^th terms of an A.P. is 24 and the sum of the 6^th and 10^th terms is 44. Find the first three terms of the A.P.
(b) Solve for x: 1 / (x + 4) − 1 / (x − 7) = 11 / 30, where x ≠ −4, 7.

Q27. CASE STUDY BASED QUESTION: [5]

Read the passage given below and answer the questions that follow:

A group of students of Class X visited a lighthouse near a coast for an educational survey. From the top of a 75 m high lighthouse from the sea level, a student observes two ships approaching it from the same direction. The angles of depression of the two ships are observed to be 30° and 45° respectively.

Sub-questions:
(a) Draw a labeled geometric diagram representing the given situation. [1]
(b) Find the distance of the nearer ship from the foot of the lighthouse. [1]
(c) Calculate the exact distance between the two ships. (Take √3 = 1.732) [3]

ANSWER KEY & DETAILED MARKING SCHEME

SECTION A: MCQs

Q. No.	Correct Answer	Q. No.	Correct Answer
Q1	(B) xy2	Q6	(A) 2.1 cm
Q2	(B) −10	Q7	(C) 10 units
Q3	(D) no solution	Q8	(A) √2 − 1
Q4	(A) ±0.2	Q9	(A) 10√3 m
Q5	(B) 22	Q10	(B) −1

SECTION B: Fill in the Blanks

Q. No.	Answer	Q. No.	Answer
Q11	Two numbers	Q16	1
Q12	0 (Zero)	Q17	30°
Q13	a + (n − 1)d	Q18	15/4
Q14	Equilateral	Q19	3
Q15	(0, y)	Q20	2√(a2 + b2)

SECTION C: Short Answers (2 Marks Each)

Q21: Let 5 − √3 = r (rational). Then √3 = 5 − r. Since r is rational, 5 − r is rational, making √3 rational, which contradicts the fact that √3 is irrational. Hence, 5 − √3 is irrational. (2 marks)

Q22: 6x² − 7x − 3 = 0 ⇒ (2x − 3)(3x + 1) = 0. Zeroes are x = 3/2 and x = −1/3. Sum of zeroes = 3/2 − 1/3 = 7/6 = −b/a; Product = (3/2)(−1/3) = −1/2 = c/a. (2 marks)

Q23: Here a = 21, d = 18 − 21 = −3. a_n = −81 ⇒ 21 + (n − 1)(−3) = −81 ⇒ (n − 1)(−3) = −102 ⇒ n − 1 = 34 ⇒ n = 35. So, 35^th term is −81. (2 marks)

Q24: Let ratio be k : 1. Point on y-axis has x-coordinate = 0. Using section formula: x = [k(−1) + 1(5)] / (k + 1) = 0 ⇒ −k + 5 = 0 ⇒ k = 5. Ratio is 5 : 1. (2 marks)

Q25: LHS = (1/cos A + sin A/cos A)(1 − sin A) = [(1 + sin A)/cos A](1 − sin A) = (1 − sin² A)/cos A = cos² A / cos A = cos A = RHS. Hence proved. (2 marks)

SECTION D: Long Answer & Case Study (5 Marks Each)

Q26:
(a) a₄ + a₈ = 24 ⇒ 2a + 10d = 24 ⇒ a + 5d = 12. Also, a₆ + a₁₀ = 44 ⇒ 2a + 14d = 44 ⇒ a + 7d = 22. Subtracting gives 2d = 10 ⇒ d = 5, a = −13. First three terms are −13, −8, −3. (2.5 marks)
(b) [(x − 7) − (x + 4)] / [(x + 4)(x − 7)] = 11/30 ⇒ −11 / (x² − 3x − 28) = 11/30 ⇒ x² − 3x − 28 = −30 ⇒ x² − 3x + 2 = 0 ⇒ (x − 1)(x − 2) = 0 ⇒ x = 1 or x = 2. (2.5 marks)

Q27:
(a) Right triangle ABD with height AB = 75 m, ship C at 45° angle (∠ACB = 45°), ship D at 30° angle (∠ADB = 30°). (1 mark)
(b) In ΔABC, tan 45° = AB / BC ⇒ 1 = 75 / BC ⇒ BC = 75 m. (1 mark)
(c) In ΔABD, tan 30° = AB / BD ⇒ 1/√3 = 75 / BD ⇒ BD = 75√3 m.
Distance between ships CD = BD − BC = 75√3 − 75 = 75(√3 − 1) = 75(1.732 − 1) = 75 × 0.732 = 54.9 m. (3 marks)