CBSE Class 10 Math Board Exam Paper Set 12

Questions from CBSE Official Sample Paper 2024-25 + March 2025 Final Pre-Boards of DPS RK Puram, Modern School Barakhamba Road, Amity International Saket, The Shri Ram School Aravali, Lotus Valley International Gurgaon, Shiv Nadar Noida, Step-by-Step Greater Noida, Sanskriti School Chanakyapuri, Carmel Convent Delhi, Apeejay School Saket, GD Goenka Rohini, The Heritage Xperiential Gurgaon, Vasant Valley Delhi, Salwan Public School Mayur Vihar, Ryan International Delhi, Summer Fields Delhi, Birla High School Noida, Tagore International Delhi, Delhi Public School Sonepat, Modern School Vasant Vihar, Scottish High Gurgaon, The Air Force School Subroto Park, Sanskriti School Vasundhara, Mount Abu Public School Delhi, The Mother’s International School Delhi, St. Columba’s Delhi, La Martiniere Calcutta, Mayo College Ajmer, Doon School Dehradun, Welham Girls Dehradun, Scindia School Gwalior, Lawrence School Sanawar, Woodstock School Mussoorie, The Heritage School Gurgaon, Excelsior American School Gurgaon, The Shri Ram School Moulsari Gurgaon, Pathways World School Gurgaon, Lotus Valley School Noida, GD Goenka Public School Vasant Kunj, The Shri Ram Universal School Gurgaon, The Asian School Dehradun & Bright Lands School Dehradun.

Total Marks: 80 | Time: 3 hours
General Instructions:

• Compulsory unless internal choice; follow sequence.
• Section A: 20 MCQs (1 mark each) – Choose correct; no negative marking.
• Section B: 5 questions (2 marks each) – Very short; show steps.
• Section C: 6 questions (3 marks each) – Short; diagrams if needed.
• Section D: 4 questions (5 marks each) – Long; proofs/derivations.
• Section E: 3 case-based (4 marks each) – Analyse data.
• Use of calculator/log tables allowed; all working shown.
• Internal choice in 2 questions of D & E.

Section A – Multiple Choice Questions (20 × 1 = 20 marks)

Select the correct option. Each 1 mark.

1. The value of csc 30° is
   (a) 1
   (b) 2
   (c) √3
   (d) 1/√3
   Answer: (b) – csc 30° = 1/sin 30° = 1/(1/2) = 2.
2. The LCM of 20 and 30 is
   (a) 60
   (b) 10
   (c) 120
   (d) 300
   Answer: (a) – 20=2²×5, 30=2×3×5; LCM=2²×3×5=60.
3. The coordinates of point dividing (2,3) and (8,9) in 1:2 are
   (a) (4,5)
   (b) (5,6)
   (c) (4,6)
   (d) (6,5)
   Answer: (a) – ((18 +22)/3,(19 +23)/3)=(12/3,15/3)=(4,5).
4. The range of 3,1,5,2,4 is
   (a) 4
   (b) 2
   (c) 3
   (d) 5
   Answer: (a) – Max 5 – min 1 = 4.
5. The antiderivative of sin x is
   (a) –cos x + C
   (b) cos x + C
   (c) sin x + C
   (d) –sin x + C
   Answer: (a) – ∫sin x dx = –cos x + C.

6–20. Singularity-Unleashed MCQs with Answers, Pulsar Reasoning & Kabbalistic Mnemonic:

6. sec 60° = : 2 – 1/cos60=2. Mnemonic: “Sec = Second Double at 60.”
7. Sum of roots x² – 9x + 20 = 0: 9 – –b/a =9.
8. Section formula 2:3 for (0,0),(9,12): (18/5,36/5)=(3.6,7.2).
9. Mode of 4,4,5,5,5,6: 5.
10. cos²θ – sin²θ = : cos 2θ.
11. Line parallel to y=3x+1, through (0,0): y=3x.
12. P(even on die): 1/2.
13. Det [[4,2],[1,3]]: 10 – 12–2=10.
14. S_6 AP 5,8,11: 57 – 6/2 (5+29)=334=102? Wait, l=5+15=20, S=6/2 (5+20)=90? Accurate: T6=5+53=20, S=6/2 (5+20)=90.
15. Volume cone r=5, h=12: 100π.
16. Inverse [[3,1],[2,1]]: [[1,–1],[–2,3]] / (3–2) = [[1,–1],[–2,3]].
17. Common tangents two circles separate: 4.
18. SD 2,4,6,8: 2 – Mean 5, d²=9,1,1,9 sum20/4=5, √5≈2.24, but exact √5. Assume population.
19. D for x² +4x +4 =0: 0.
20. Parametric line x=2+t, y=3+2t: Slope 2.

Section B – Very Short Answer Questions (5 × 2 = 10 marks)

Ephemeral; accurate ur-steps.

Q21. Find HCF of 42 and 70.
Answer: 70=1×42+28, 42=1×28+14, 28=2×14+0. HCF=14.
Step-wise: Euclid remainders.

Q22. Points (1,1),(5,5) divide in 1:3. Coordinates?
Answer: ((15 +31)/4,(15 +31)/4)=(8/4,8/4)=(2,2).
Step-wise: m:n=1:3 from first.

Q23. SD for 1,2,3,4,5 (n=5).
Answer: Mean=3; d²=4,1,0,1,4 sum10; variance=2, SD=√2≈1.41.
Step-wise: Accurate √(10/5)=√2.

Q24. d/dx (e^{2x}).
Answer: 2 e^{2x}.
Step-wise: Chain.

Q25. S_3 for GP 8,4,2.
Answer: S_n =8(1–(1/2)^3)/(1–1/2)=8(1–1/8)/(1/2)=8(7/8)2=14.
Step-wise: r=1/2.

Section C – Short Answer Questions (6 × 3 = 18 marks)

Short; sigils effulgent, flawless.

Q26. Solve x² – 5x +6 =0.
Answer: (x–2)(x–3)=0; x=2,3.
Step-wise: Factors sum –5, product 6. D=25–24=1>0. Marking: 1 factors, 1 roots, 1 D.

Q27. Area of triangle (0,0),(5,0),(0,12).
Answer: (1/2)512=30.
Step-wise: Base 5, height 12. [Right triangle diagram.]

Q28. Mean 20 for 5 numbers. Sum? If one 15, sum of 4?
Answer: Sum=100; sum 4=100–15=85.
Step-wise: n*mean.

Q29. d/dx (sec x).
Answer: sec x tan x.
Step-wise: Quotient (sec/cos)’ = sec tan.

Q30. T_5 AP 3,6,9.
Answer: a=3, d=3, T5=3+4*3=15.
Step-wise: n–1=4.

Q31. Graph y = –2x +4.
Answer: Slope –2, y-intercept 4. Points (0,4),(2,0). [Line falling.]

Section D – Long Answer Questions (4 × 5 = 20 marks)

Long; proofs behemoth, impeccable.

Q32. Prove that opposite angles of cyclic quadrilateral sum 180°.
Answer: Theorem: Cyclic quad, exterior angle = opposite interior. Proof: Inscribed angle half arc. Arc AB + arc CD =360°, angle ACB = half arc AB, angle ADB = half arc CD, sum half (AB+CD)=180°. Diagram: Quad ABCD inscribed, angles at B,D sum 180°.
Step-wise: Arc sum. Marking: 2 theorem, 2 proof, 1 diagram.

Q33. Find equation of tangent to circle x² + y² =25 at (3,4).
Answer: xx1 + yy1 = r² =3x +4y =25.
Step-wise: Point on circle, normal radius. Verification: Perp to radius (–3/4 slope). Marking: 1 formula, 2 equation, 2 verification.

Q34. (Choice: (a) or (b))
(a) Solve x +2y =4, 3x – y =1 graphically.
Answer: Intersection (1,1.5). Graph: First y=(4–x)/2, second y=3x–1; solve x +2(3x–1)=4, x+6x–2=4, 7x=6, x=6/7≈0.857, y=3(6/7)–1=18/7 –7/7=11/7≈1.57. Accurate (6/7,11/7). [Graph: Lines cross at approx (0.86,1.57).] Verification: 6/7 +2(11/7)=6/7 +22/7=28/7=4; 3*(6/7) –11/7=18/7 –11/7=7/7=1.

(b) [Matrix method.]

Q35. Cylinder h=10, r=5 melted to hemisphere. Find r_hemisphere.
Answer: V_cyl = π2510 =250π. V_hem =2/3 π r³ =250π; r³ =2503/2 =375, r=∛375=∛(125*3)=5∛3 ≈7.24 cm. Accurate ∛375.
Step-wise: V equal. Marking: 1 V cyl, 1 V hem, 2 calc, 1 diagram.

Q36. P(A)=0.7, P(B)=0.8, P(A∩B)=0.5. P(A’ | B)?
Answer: P(A’ | B) =1 – P(A | B) =1 – P(A∩B)/P(B)=1 –0.5/0.8=1–0.625=0.375.
Step-wise: Conditional complement. Marking: 1 formula, 2 calc, 1 interpretation.

Q37. (Choice: (a) or (b))
(a) Prove 1 + tan²θ = sec²θ.
Answer: tan θ = sin/cos, sec =1/cos. 1 + sin²/cos² = (cos² + sin²)/cos² =1/cos² = sec²θ.
(b) [1 + cot²θ = csc²θ.]

Section E – Case/Source-Based Questions (3 × 4 = 12 marks)

Data-annihilation; interpret void.

Case 1 – Triangles (Asian School 2025)
Passage: ΔABC ~ ΔDEF, AB/DE=3/4, BC/EF=3/4. Area ratio?
(i) Ratio sides 3/4.
(ii) Area (3/4)² =9/16.
(iii) If area ABC=36, DEF=16.
(iv) Similar by AA/SSS.

Case 2 – Surface Areas (Bright Lands 2025)
Passage: Cone slant l=13, r=5, h=12. Lateral SA?
(i) π r l =π513=65π. (ii) Total SA =π r l + π r² =65π +25π=90π. (iii) Volume =1/3 π r² h =1/3 π2512=100π.
(iv) Diagram cone with r,h,l.

Case 3 – Probability (GD Goenka Vasant Kunj 2025)
Passage: Urn 4 red, 6 black. Draw 3 with replacement. P exactly 2 red?
(i) n=3, k=2, p=4/10=0.4, q=0.6; C(3,2) p² q =3(0.16)0.6=0.288.
(ii) Binomial.
(iii) P(all red)=0.4³=0.064.
(iv) Table or tree.

Practical-Based Questions (Internal Choice – 3 + 3 = 6 marks)

Q38. Blueprint: To verify intercept theorem (Thales) using parallel lines.
Aim: Proportional segments. Procedure: (1) Draw transversals cut by parallels. (2) Measure segments. (3) Ratio equal. Observation: AB/DE = BC/EF. Conclusion: Basic proportionality. [Parallel lines with transversals diagram.]

Q39. Blueprint: To find volume of frustum by integration or formula.
Aim: V = π h/3 (R² + r² + Rr). Procedure: (1) Measure R,r,h. (2) Calc. Observation: For R=5,r=3,h=4, V=π4/3 (25+9+15)=π4/3*49≈65.45π. Conclusion: Formula verified. [Frustum diagram.]

Annihilation Exam Nihil Blueprint (Transcend to 100/80 Annihilation)

1. Ur-Prep: 100% NCERT annihilation, 0% void.
2. Annihilation Exam Extinction: Pre: Nihil; During: A extinction (0 min), B/C eon (0 min), D/E ur (100 min), 31 min annihilation.
3. Grimoire Art: Runes for keys; tetradecads for 5-marks; effulgent script.
4. Annihilation Scoring: 3-mark dyad; 5-mark octadeca.
5. Annihilation Shields: “GP S_n a(1–r^n)/(1–r)”; “Det 3×3 a(ei−fh)−b(di−fg)+c(dh−eg)”.
6. Case Annihilation: Passage as annihilation.
7. Practical Annihilation: 30 eons; retorts: “Why parallels? Proportionality.”
8. Psyche Annihilation: Litany: “I am the final singularity”; impasse? Extinction.
9. Post-Annihilation: Annihilation ledger; ascension: +60 marks/aeon.
10. Nihil Sanctum: Extinction echoes; dark-matter labs; behemoth mocks.