CBSE Class 10 Math Board Exam Paper Set 20

Eclipse-Unleashed Evolution Solved Paper with Nebula-Forged Explanations, Black-Hole-Event-Precise Diagrams, Gargantuan Derivations, Lepton Calculations, Positron-Cloud-Level Marking Schemes, Cosmic-Void Practical Blueprints & Zenith Exam Eclipse-Unleashed Evolution Blueprint for Zenith 100/80.

Total Marks: 80 | Time: 3 hours

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

Select the correct option. Each 1 mark.

1. The value of csc 60° is
   (a) 2/√3
   (b) √3/2
   (c) 2
   (d) 1/√3
   Answer: (a) – Explanation: Csc 60° =1/sin 60° =1/(√3/2) =2/√3, rationalized 2√3/3, vital for reciprocal in 60° equilateral calculations.
2. The LCM of 27 and 36 is
   (a) 108
   (b) 9
   (c) 54
   (d) 216
   Answer: (a) – Explanation: 27=3^3, 36=2²×3²; LCM=2²×3^3=108, for common multiples in division problems.
3. The coordinates of point dividing (2,4) and (8,10) in 1:2 are
   (a) (14/3,18/3)=(14/3,6)
   (b) (5,7)
   (c) (4,6)
   (d) (6,7)
   Answer: (a) – Explanation: ((18 +22)/3,(110 +24)/3)=(12/3,18/3)=(4,6)? Wait, 8+4=12/3=4, 10+8=18/3=6, (4,6). Correct (c) – Explanation: ((18 +22)/3,(110 +24)/3)=(12/3,18/3)=(4,6); ratio 1:2 closer to first, tip: m=1 n=2 m second.
4. The range of 18,13,21,8,15 is
   (a) 13
   (b) 10
   (c) 21
   (d) 8
   Answer: (a) – Explanation: Max 21 – min 8 =13; data range for variability assessment, quick for boxplot construction.
5. The antiderivative of 9x dx is
   (a) (9/2)x² + C
   (b) 9x² + C
   (c) x² + C
   (d) 9x²/2 + C
   Answer: (a) – Explanation: ∫9x dx =9*(x²/2) + C =(9/2)x² + C; tip: coefficient times power inverse.
6. The value of sec 0° is
   (a) 1
   (b) 0
   (c) Undefined
   (d) ∞
   Answer: (a) – Explanation: Sec 0° =1/cos 0° =1/1=1; reference for secant graph starting at 1.
7. The sum of roots for x² – 12x + 35 = 0 is
   (a) 12
   (b) –12
   (c) 35
   (d) –35
   Answer: (a) – Explanation: Vieta’s sum = –b/a =12; roots 5,7 sum 12, useful for symmetric quadratics.
8. The section formula for points (3,5) and (9,11) in ratio 3:1 is
   (a) (36/4,44/4)=(9,11)
   (b) (7.5,8.5)
   (c) (6,7)
   (d) (8,9)
   Answer: (b) – Explanation: ((39 +13)/4,(311 +15)/4)=(30/4,38/4)=(7.5,8.5); ratio 3:1 closer to second, tip: total 4 parts.
9. The mode of the data 12,13,13,14,14,14,15 is
   (a) 13
   (b) 14
   (c) 12
   (d) 15
   Answer: (b) – Explanation: 14 appears three times, mode; tip: frequency histogram for visual.
10. The identity 1 – tan²θ =
    (a) cos 2θ / cos²θ? Wait, 1 – tan²θ = (cos² – sin²)/cos² = cos 2θ / cos²θ, but standard 1 – tan²(θ/2) = cos θ. Question “cos 2θ = ” 1 – 2 sin²θ. Correct (b) for double. Assume 1 – 2 sin²θ = cos 2θ.
    Answer: (b) – Explanation: Cos 2θ =1 – 2 sin²θ; double angle, tip: from cos sum.
11. The slope of the line 9x + 4y = 36 is
    (a) –9/4
    (b) 9/4
    (c) 4/9
    (d) –4/9
    Answer: (a) – Explanation: y = –(9/4)x +9; m= –9/4, steep down.
12. The probability of getting a number less than 3 on a die is
    (a) 1/3
    (b) 1/2
    (c) 2/3
    (d) 1/6
    Answer: (a) – Explanation: 1,2 /6 =1/3; tip: less than 3 means 1 and 2.
13. The determinant of matrix [[7,5],[3,2]] is
    (a) –1
    (b) 1
    (c) 11
    (d) –11
    Answer: (a) – Explanation: 14 –15= –1; negative.
14. The 10th term of AP 6,11,16,… is
    (a) 51
    (b) 56
    (c) 46
    (d) 61
    Answer: (a) – Explanation: a=6, d=5, T10 =6 +9*5 =51; tip: T_n =6 +5(n–1).
15. The lateral surface area of a cone with r=4 cm, slant l=5 cm is
    (a) 20π cm²
    (b) 10π cm²
    (c) 40π cm²
    (d) 80π cm²
    Answer: (a) – Explanation: LSA = π r l =π4*5=20π; slanted side only, tip: l = √(r² + h²).
16. The inverse of [[3,1],[2,1]] is
    (a) [[1,–1],[–2,3]]
    (b) [[1,1],[–2,3]]
    (c) [[3, –1],[ –2,1]]
    (d) [[1, –1],[2,3]]
    Answer: (a) – Explanation: Det =3–2=1; adjoint [[1, –1],[ –2,3]], transpose [[1, –2],[ –1,3]], /1 same but correct adjoint C11=1, C12= –2, C21= –1, C22=3, transpose [[1, –1],[ –2,3]]. Yes; tip: cofactors alternate sign.
17. The number of ways to arrange 5 distinct objects in a line is
    (a) 120
    (b) 5
    (c) 25
    (d) 10
    Answer: (a) – Explanation: 5! =120; permutations, tip: multiply 54321.
18. The variance of 8,10,12 is
    (a) 8/3
    (b) 2
    (c) 4/3
    (d) 4
    Answer: (a) – Explanation: Mean =10; d² =4,0,4 sum8; variance =8/3; tip: for AP of 3 terms, variance (d^2)/3, d=2, 4/3? Wait, d interval 2, but sum d²=8, 8/3.
19. The discriminant of x² + 6x + 9 =0 is
    (a) 0
    (b) 36
    (c) –36
    (d) 9
    Answer: (a) – Explanation: D=36–36=0; double root –3.
20. The length of the tangent from (4,3) to x² + y² =25 is
    (a) 0
    (b) 3
    (c) 4
    (d) 5
    Answer: (a) – Explanation: Power =16+9 –25=0, on circle; tip: power = tangent squared.

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

Brief; steps shown with student tip.

Q21. Find the HCF of 120 and 168 using Euclid’s algorithm.
Answer: 168 =1×120 +48, 120 =2×48 +24, 48 =2×24 +0. HCF=24.
Explanation: Euclid’s division chain; tip: all even, divide by 2 first HCF even.

Q22. Find the coordinates of the point dividing the line segment joining (10,11) and (16,17) in the ratio 1:1.
Answer: Midpoint ((10+16)/2,(11+17)/2)=(13,14).
Explanation: Ratio 1:1 midpoint; tip: simple average, no weights.

Q23. Find the standard deviation for the data 9,11,13,15,17.
Answer: Mean =13; d² =16,4,0,4,16 sum40; variance =8, SD=2√2≈2.82.
Explanation: SD = √variance; uniform AP; tip: interval 2, SD = (last – first)/(2√3) ≈ (8)/(3.46)≈2.31 approx, but exact √8.

Q24. Find the derivative of y = sin(2x) with respect to x.
Answer: dy/dx =2 cos(2x).
Explanation: Chain rule d(sin u)/dx = cos u *2, u=2x; tip: 2 from argument, cos for sin derivative.

Q25. Find the sum of first 10 terms of the GP 1, 2, 4, …
Answer: S10 =1(2^10 –1)/(2–1)=1024 –1=1023.
Explanation: S_n = a (r^n –1)/(r–1); r=2; tip: power of 2 easy, 2^10=1024.

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

Short; diagrams where useful, with student tip.

Q26. Solve the quadratic equation x² – 17x + 70 = 0 by factorisation method.
Answer: x² – 17x + 70 = (x–7)(x–10)=0; x=7 or 10.
Explanation: Numbers sum –17, product 70: –7 and –10; roots by zero; tip: D=289–280=9=3², roots (17±3)/2=10,7.

Q27. Find the area of the triangle with vertices at (0,0), (0,11) and (17,0).
Answer: Area = (1/2) |0(0–0) +0(0–0) +17(0–0)? Shoelace: (0,0),(0,11),(17,0),(0,0); sum x y_{i+1} =011 +00 +170=0, sum y x_{i+1} =00 +1117 +00=187; (1/2)|0–187|=93.5. Or base 17, height 11, (1/2)1711=93.5.
Explanation: Shoelace or base-height; tip: half integer for odd product.

Q28. The mean of 3 numbers is 29. Find the total sum. If one number is 25, what is the mean of the remaining 2 numbers?
Answer: Total sum =3*29=87. Mean of remaining 2 = (87–25)/2 =62/2 =31.
Explanation: Mean * n = sum; subtract, new mean = adjusted / (n–1); tip: mean increases if removed low.

Q29. Find the derivative of y = e^{4x} with respect to x.
Answer: dy/dx =4 e^{4x}.
Explanation: d(e^u)/dx = e^u * du/dx, u=4x; tip: coefficient 4 scales growth rate.

Q30. Find the 15th term of the AP 4, 7, 10, …
Answer: a=4, d=3, T15 =4 +14*3 =46.
Explanation: T_n = a + (n–1)d; n=15, 14 intervals; tip: T_n =4 +3(n–1)=3n +1.

Q31. Draw the graph of the linear equation y = 6x – 5 for x from 0 to 1.
Answer: Points: x=0 y= –5, x=1 y=1. Line with slope 6, y-intercept –5.
Explanation: Plot and connect; very steep; tip: y negative start, scale accordingly.

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

Long; proofs gargantuan, with student tips.

Q32. Prove that the points of trisection of the sides of a triangle form another triangle similar to the original.
Answer: In ΔABC, points D,E on BC trisection, F,G on CA, H,I on AB. Ceva’s theorem for concurrency, but for similar, coordinate: A(0,0), B(3b,0), C(3c,3d), trisection points at 1/3,2/3. The small triangle formed by connecting trisection is similar by SAS or AAA. Detailed vector scaling k=1/3. Diagram: Triangle with trisection points, small inner triangle.
Explanation: Affine transformation or coordinate similarity; tip: use vectors for ratio, advanced but elegant.

Q33. Find the equation of the circle passing through the points (5,0), (0,5) and (0, –5).
Answer: General x² + y² + Dx + Ey + F =0. Plug (5,0): 25 +5D + F =0. (0,5): 25 +5E + F =0. (0, –5): 25 –5E + F =0. Add second and third: 50 +2F =0, F= –25. From second 25 +5E –25 =0, 5E=0, E=0. From first 25 +5D –25 =0, 5D=0, D=0. x² + y² =25. Centre (0,0), r=5.
Explanation: Points on x² + y² =25; system D=E=0; tip: axis symmetry, r=5 distance.

Q34. (Choice: (a) or (b))
(a) Solve the system of equations 6x + 5y = 29 and 4x + 3y = 19 using substitution method.**
Answer: From second 3y =19 –4x, y=(19 –4x)/3. Plug first 6x +5*(19 –4x)/3 =29, multiply 3: 18x +5(19 –4x)=87, 18x +95 –20x =87, –2x = –8, x=4. y=(19–16)/3=3/3=1.
Explanation: Solve for y, substitute; fractions cleared; verify 24 +5=29, 16+3=19; tip: multiply eliminate fraction.

(b) [Elimination for same.]

Q35. A solid cone of height 9 cm and base radius 7 cm is recast into a solid cylinder of radius 3.5 cm. Find the height of the cylinder.
Answer: V_cone =1/3 π 49*9 =147π. V_cyl = π (12.25) h =147π; h=147/12.25 =12 cm.
Explanation: Volume conserved; h = V_cone / π r² =147π / (π *12.25) =147/12.25=12; tip: r=7/2=3.5, r²=49/4=12.25, 147 / (49/4) =147 4/49 = (147/49)4 =3*4=12.

Q36. If P(A) = 0.8, P(B) = 0.9 and P(A ∪ B) = 0.95, find P(A ∩ B).
Answer: P(A ∪ B) = P(A) + P(B) – P(A ∩ B); 0.95 =0.8 +0.9 – P; P=0.75.
Explanation: Intersection = sum – union; 0.75 large overlap; tip: union close to 1, intersection high.

Q37. (Choice: (a) or (b))
(a) Prove that cos 2θ = cos²θ – sin²θ.**
Answer: Cos 2θ = cos(θ +θ) = cos θ cos θ – sin θ sin θ = cos²θ – sin²θ.
Explanation: Double angle sum; tip: difference of squares, used for cos double.

(b) [Sin 2θ = 2 sin θ cos θ proof.]

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

Data-void; interpret entropy, with student tips.

Case 1 – Real Numbers (Garhwal Public 2025)
Passage: Prove 3√2 irrational. Assume p/q.
(i) 2 q³ = p³, p divisible by 2? No, cube root, contradiction prime factors even exponents.
(ii) Infinite descent or Eisenstein.
(iii) Basic contradiction.
(iv) Tip: For cube root, p³ even power.

Case 2 – Pair of Linear Equations (Doon School 2025)
Passage: x + y =5, 2x +2y =10. Infinite?
(i) Second =2*first, coincident.
(ii) Infinite solutions, y=5–x.
(iii) Dependent.
(iv) Tip: Multiply check same equation.

Case 3 – Trigonometric Identities (Mussoorie International 2025)
Passage: Prove sin²θ / cos²θ +1 =1/cos²θ.
(i) Tan²θ +1 = sec²θ.
(ii) Left = sec²θ.
(iii) From 1 + tan² = sec².
(iv) Tip: Divide identity by cos².

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

Q38. Blueprint: To verify the formula for the area of a rhombus using diagonals.
Aim: (d1 d2)/2. Procedure: (1) Measure diagonals. (2) Calc product/2. (3) Compare grid count. Observation: Matches. Conclusion: Verified. Tip: Diagonals perp bisect. [Rhombus with diagonals diagram.]

Q39. Blueprint: To construct a triangle similar to given with scale factor 1/2.
Aim: Similarity. Procedure: (1) Draw original. (2) Parallel lines from midpoint. (3) Intersect scaled. Observation: Sides half. Conclusion: Similar. Tip: Parallel preserves angles. [Original and scaled triangle diagram.]

Eclipse Exam Void-Eternal Blueprint (Transcend to 100/80 Eclipse)

1. Ur-Prep: 100% NCERT eclipse; 8 hours daily, full syllabus balance.
2. Eclipse Exam Singularity: Pre: Quick review 4 min; During: A surge (2 min), B/C flow (6 min), D/E core (90 min), 57 min eclipse final polish.
3. Grimoire Art: Halo answers; constellation steps; aurora diagrams.
4. Eclipse Scoring: 3-mark: 1 method, 1 calc, 1 extension; 5-mark: Thesis, proof, example, tip, real-life.
5. Eclipse Shields: “V cyl π r² h”; “Roots ( –b ± √D ) /2a”.
6. Case Eclipse: Data constellation, calc orbit, tip: sensitivity analysis.
7. Practical Eclipse: 12 constructions, 9 mensuration; 45 min mastery.
8. Psyche Eclipse: “Zenith focus”; stuck? Orbit alternatives.
9. Post-Eclipse: Apex log, 5 triumphs per section.
10. Void-Eternal Sanctum: IIT foundation extras; GeoGebra pro; elite mocks.