X Mat SP2

CBSE CLASS X MATHEMATICS

MODEL QUESTION PAPER - 2

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 LCM(91, 26) = 182, then HCF(91, 26) is equal to: [1]

(A) 13

(B) 26

(C) 7

(D) 91

Q2. The number of zeroes for a polynomial p(x) whose graph intersects the x-axis at 3 distinct points is: [1]

(A) 1

(B) 2

(C) 3

(D) 4

Q3. For what value of k do the equations 3x − y + 8 = 0 and 6x − ky = −16 represent coincident lines? [1]

(A) 1/2

(B) −1/2

(C) 2

(D) −2

Q4. The discriminant of the quadratic equation 2x² − 4x + 3 = 0 is: [1]

(A) −8

(B) 10

(C) −16

(D) 8

Q5. What is the common difference of an A.P. in which a₁₈ − a₁₄ = 32? [1]

(A) 8

(B) −8

(C) 4

(D) −4

Q6. Two similar triangles ΔABC and ΔPQR have perimeters 36 cm and 24 cm respectively. If PQ = 10 cm, then AB is equal to: [1]

(A) 15 cm

(B) 20 cm

(C) 12 cm

(D) 18 cm

Q7. The perimeter of a triangle with vertices (0, 4), (0, 0), and (3, 0) is: [1]

(A) 5 units

(B) 12 units

(C) 11 units

(D) 7 + √5 units

Q8. If sin A = 1/2, then the value of (3 cos A − 4 cos³ A) is: [1]

(A) 1

(B) 0

(C) 1/2

(D) 1 / √2

Q9. A ladder 15 m long leans against a wall making an angle of 60° with the wall. The height of the point where the ladder touches the wall is: [1]

(A) 7.5 m

(B) 15√3 m

(C) 7.5√3 m

(D) 15 m

Q10. If the distance between points A(x, −1) and B(5, 3) is 5 units, then the value of x is: [1]

(A) 2 or 8

(B) −2 or 8

(C) 2 or −8

(D) −2 or −8

SECTION B: FILL IN THE BLANKS [10 MARKS]

Q11. Every composite number can be uniquely expressed as a product of ____________________ numbers. [1]

Q12. If the sum of the zeroes of quadratic polynomial ax² + bx + c is 0, then coefficient b is equal to ____________________. [1]

Q13. If 2x, x + 10, and 3x + 2 are in A.P., then the value of x is ____________________. [1]

Q14. If a line divides any two sides of a triangle in the same ratio, then the line is ____________________ to the third side. [1]

Q15. The point equidistant from points A(−2, 0) and B(2, 0) on the y-axis is ____________________. [1]

Q16. Value of (1 + tan² θ)(1 − sin θ)(1 + sin θ) is equal to ____________________. [1]

Q17. If the height of a tower is equal to the length of its shadow, the angle of elevation of the sun is ____________________. [1]

Q18. If the pair of linear equations has a unique solution, the lines represented by them are ____________________ lines. [1]

Q19. If a and b are roots of x² − 5x + 6 = 0, then value of (a + b) is ____________________. [1]

Q20. The distance of point P(3, 4) from the x-axis is ____________________ units. [1]

SECTION C: SHORT ANSWER QUESTIONS [10 MARKS]

Q21. Explain why 7 × 11 × 13 + 13 is a composite number. [2]

Q22. Solve the system of linear equations by substitution method: 2x + 3y = 11 and 2x − 4y = −24. [2]

Q23. Find the 20^th term from the last term of the A.P. 3, 8, 13, ..., 253. [2]

Q24. Find the point on the x-axis which is equidistant from (2, −5) and (−2, 9). [2]

Q25. Prove that: (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A. [2]

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

Q26. [5]
(a) Find two consecutive odd positive integers, sum of whose squares is 290.
(b) State and prove Basic Proportionality Theorem (Thales Theorem).

Q27. CASE STUDY BASED QUESTION: [5]

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

A direct measurement of distance across a lake was not possible due to water obstruction. An engineering team set up surveying instruments at point P on one bank. Two landmarks A and B were identified on opposite sides across the water. Point P was chosen such that ∠APB = 90°. From point P, landmark A is at a distance of 60 m and landmark B is at a distance of 80 m. An observer at P tracks a drone flying directly above landmark A at an angle of elevation of 45°.

Sub-questions:
(a) Calculate the direct straight-line distance AB across the lake between the two landmarks. [2]
(b) Find the height at which the drone is flying directly above landmark A. [1]
(c) If the drone moves directly above landmark B keeping its height unchanged, find the new angle of elevation of the drone from point P. [2]

ANSWER KEY & DETAILED MARKING SCHEME

SECTION A: MCQs

Q. No.	Correct Answer	Q. No.	Correct Answer
Q1	(A) 13	Q6	(A) 15 cm
Q2	(C) 3	Q7	(B) 12 units
Q3	(C) 2	Q8	(B) 0
Q4	(A) −8	Q9	(A) 7.5 m
Q5	(A) 8	Q10	(A) 2 or 8

SECTION B: Fill in the Blanks

Q. No.	Answer	Q. No.	Answer
Q11	Prime	Q16	1
Q12	0 (Zero)	Q17	45°
Q13	6	Q18	Intersecting
Q14	Parallel	Q19	5
Q15	(0, 0) / Origin	Q20	4

SECTION C: Short Answers (2 Marks Each)

Q21: 7 × 11 × 13 + 13 = 13 × (7 × 11 + 1) = 13 × (77 + 1) = 13 × 78 = 13 × 13 × 6. Since it has prime factors other than 1 and itself, it is a composite number. (2 marks)

Q22: 2x + 3y = 11 ...(i) and 2x − 4y = −24 ...(ii). From (i), 2x = 11 − 3y. Substitute in (ii): (11 − 3y) − 4y = −24 ⇒ −7y = −35 ⇒ y = 5. Then 2x = 11 − 15 = −4 ⇒ x = −2. (2 marks)

Q23: Reversed A.P.: 253, 248, ..., 13, 8, 3 with first term a = 253 and d = −5. 20^th term a₂₀ = 253 + (20 − 1)(−5) = 253 − 95 = 158. (2 marks)

Q24: Let point be P(x, 0). Equidistant from A(2, −5) and B(−2, 9) ⇒ PA² = PB². So, (x − 2)² + (0 + 5)² = (x + 2)² + (0 − 9)² ⇒ x² − 4x + 4 + 25 = x² + 4x + 4 + 81 ⇒ −8x = 56 ⇒ x = −7. Point is (−7, 0). (2 marks)

Q25: LHS = sin² A + cosec² A + 2 + cos² A + sec² A + 2 = (sin² A + cos² A) + 4 + (1 + cot² A) + (1 + tan² A) = 1 + 4 + 1 + cot² A + 1 + tan² A = 7 + tan² A + cot² A = RHS. Hence proved. (2 marks)

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

Q26:
(a) Let numbers be x and x + 2. Equation: x² + (x + 2)² = 290 ⇒ 2x² + 4x + 4 = 290 ⇒ x² + 2x − 143 = 0 ⇒ (x + 13)(x − 11) = 0. Since positive, x = 11. Numbers are 11 and 13. (2.5 marks)
(b) Statement: If a line is drawn parallel to one side of a triangle, it divides other two sides in same ratio. Proof: Draw diagram, construct altitudes, use area ratio concept to derive AD/DB = AE/EC. (2.5 marks)

Q27:
(a) In right ΔAPB, AB² = AP² + PB² = 60² + 80² = 3600 + 6400 = 10000 ⇒ AB = 100 m. (2 marks)
(b) Let height be h. tan 45° = h / AP ⇒ 1 = h / 60 ⇒ h = 60 m. (1 mark)
(c) Drone height = 60 m above B. Angle of elevation from P: tan θ = Height / PB = 60 / 80 = 3/4 = 0.75 ⇒ θ = tan⁻¹(0.75) ≈ 36.87°. (2 marks)