CBSE Class 10 Math Board Exam Paper Set 19

Zenith-Unleashed Evolution Solved Paper with Cosmic-Nexus Explanations, Pulsar-Event-Precise Diagrams, Titanic Derivations, Hadron Calculations, Electron-Nexus-Level Marking Schemes, Nebula-Vortex Practical Blueprints & Supernova Exam Zenith-Unleashed Evolution Blueprint for Supernova 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 sec 60° is
   (a) 2
   (b) 1/2
   (c) √3
   (d) 1/√3
   Answer: (a) – Explanation: Sec 60° =1/cos 60° =1/(1/2) =2, reciprocal cosine for 60°, used in equilateral side calculations.
2. The LCM of 32 and 48 is
   (a) 96
   (b) 16
   (c) 32
   (d) 144
   Answer: (a) – Explanation: 32=2^5, 48=2^4×3; LCM=2^5×3=96, for binary powers in computer science multiples.
3. The coordinates of point dividing (6,4) and (12,10) in 1:2 are
   (a) (18/3,14/3)=(6,4.67)
   (b) (8,6)
   (c) (7,5)
   (d) (9,7)
   Answer: (a) – Explanation: ((112 +26)/3,(110 +24)/3)=(24/3,18/3)=(8,6)? Wait, 12+12=24/3=8, 10+8=18/3=6, (8,6). Correct (b) – Explanation: ((112 +26)/3,(110 +24)/3)=(24/3,18/3)=(8,6); ratio 1:2 closer to first, tip: m=1 n=2 m second.
4. The range of 20,15,23,10,17 is
   (a) 13
   (b) 10
   (c) 23
   (d) 5
   Answer: (a) – Explanation: Max 23 – min 10 =13; data extent, for range in quality control.
5. The antiderivative of 10x dx is
   (a) 5x² + C
   (b) 10x² + C
   (c) x² + C
   (d) 10x²/2 + C
   Answer: (a) – Explanation: ∫10x dx =10*(x²/2) + C =5x² + C; tip: coefficient halves with power.
6. The value of csc 0° is
   (a) Undefined
   (b) 1
   (c) 0
   (d) ∞
   Answer: (a) – Explanation: Csc 0° =1/sin 0° =1/0 undefined; asymptote in csc graph.
7. The sum of roots for x² – 13x + 42 = 0 is
   (a) 13
   (b) –13
   (c) 42
   (d) –42
   Answer: (a) – Explanation: Vieta’s sum = –b/a =13; roots 6,7 sum 13, for symmetric.
8. The section formula for points (7,8) and (13,14) in ratio 2:1 is
   (a) (39/3,42/3)=(13,14)
   (b) (11,12)
   (c) (10,11)
   (d) (12,13)
   Answer: (b) – Explanation: ((213 +17)/3,(214 +18)/3)=(33/3,36/3)=(11,12); ratio 2:1 closer to second, tip: total 3 parts.
9. The mode of the data 14,15,15,16,16,16,17 is
   (a) 15
   (b) 16
   (c) 14
   (d) 17
   Answer: (b) – Explanation: 16 appears three times, mode; tip: scan for max count.
10. The identity 1 + sin²θ =
    (a) csc²θ – cot²θ? Wait, 1 + sin² = 2 – cos², but standard 1 + tan² = sec². Assume 1 + tan²θ = sec²θ.
    Answer: (b) – Explanation: 1 + tan²θ = sec²θ; from sin²/cos² +1 =1/cos²; tip: divide by cos².
11. The slope of the line 10x + 3y = 30 is
    (a) –10/3
    (b) 10/3
    (c) 3/10
    (d) –3/10
    Answer: (a) – Explanation: y = –(10/3)x +10; m= –10/3, steep down.
12. The probability of getting a sum 7 in two dice is
    (a) 1/6
    (b) 5/36
    (c) 1/36
    (d) 6/36
    Answer: (d) – Explanation: Ways (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) =6/36=1/6; tip: list pairs for small.
13. The determinant of matrix [[8,6],[4,3]] is
    (a) 0
    (b) 24
    (c) –24
    (d) 12
    Answer: (a) – Explanation: 24 –24=0; singular, rows proportional.
14. The 11th term of AP 7,12,17,… is
    (a) 62
    (b) 67
    (c) 57
    (d) 72
    Answer: (a) – Explanation: a=7, d=5, T11 =7 +10*5 =57? Wait, 7+50=57, option (c). Correct (c) – Explanation: (11–1)=10 d=50, 7+50=57; tip: T_n =7 +5(n–1).
15. The total surface area of a cone with r=5 cm, l=13 cm is
    (a) 90π cm²
    (b) 65π cm²
    (c) 130π cm²
    (d) 25π cm²
    Answer: (a) – Explanation: LSA π r l =65π + base π25 =90π; tip: total includes base.
16. The inverse of [[4,2],[1,1]] is
    (a) [[1,–2],[–1,4]] / (4–2) = [[1,–2],[–1,4]] /2 = [[0.5,–1],[–0.5,2]]
    (b) [[1,2],[1,4]]
    (c) [[1, –2],[1,4]]
    (d) [[0.5,1],[0.5,2]]
    Answer: (a) – Explanation: Det =4–2=2; adjoint [[1, –1],[ –2,4]], transpose [[1, –2],[ –1,4]], /2 = [[0.5, –1],[ –0.5,2]]; tip: scale by 1/det.
17. The number of ways to arrange 6 distinct objects in a circle is
    (a) 120
    (b) 720
    (c) 60
    (d) 360
    Answer: (c) – Explanation: (n–1)! =5! =120 for rotations same? Wait, circular (n–1)! =120, but option (a). Correct (a) – Explanation: (6–1)! =120; tip: fix one position.
18. The variance of 9,11,13 is
    (a) 8/3
    (b) 2
    (c) 4/3
    (d) 4
    Answer: (a) – Explanation: Mean =11; d² =4,0,4 sum8; variance =8/3; tip: AP interval 2, variance (2^2)/3 =4/3? Wait, for unit variance, but 8/3 for this.
19. The discriminant of x² – 5x + 6 =0 is
    (a) 1
    (b) 25
    (c) –1
    (d) 9
    Answer: (a) – Explanation: D=25–24=1; roots 2,3.
20. The length of the tangent from (5,0) to x² + y² =16 is
    (a) 3
    (b) 0
    (c) 4
    (d) 5
    Answer: (a) – Explanation: Power =25 +0 –16=9, length √9=3; tip: power = tangent².

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

Brief; steps shown with student tip.

Q21. Find the HCF of 135 and 225 using Euclid’s algorithm.
Answer: 225 =1×135 +90, 135 =1×90 +45, 90 =2×45 +0. HCF=45.
Explanation: Euclid’s chain; tip: all multiples of 45, spot pattern.

Q22. Find the coordinates of the point dividing the line segment joining (11,12) and (17,18) in the ratio 1:1.
Answer: Midpoint ((11+17)/2,(12+18)/2)=(14,15).
Explanation: Ratio 1:1 midpoint; tip: no ratio, just average.

Q23. Find the standard deviation for the data 10,12,14,16,18.
Answer: Mean =14; d² =16,4,0,4,16 sum40; variance =8, SD=2√2≈2.82.
Explanation: SD = √variance; even AP; tip: interval 2, SD =2 * √( (5^2 –1)/12 ) ≈2.83 formula for uniform.

Q24. Find the derivative of y = sin(4x) with respect to x.
Answer: dy/dx =4 cos(4x).
Explanation: Chain rule d(sin u)/dx = cos u *4, u=4x; tip: 4 scales amplitude.

Q25. Find the sum of first 11 terms of the GP 1, 1/2, 1/4, …
Answer: S11 =1(1 – (1/2)^11)/(1 –1/2)=1(1 –1/2048)/(1/2)=(2047/2048)*2 =2047/1024≈1.999.
Explanation: S_n = a (1 – r^n)/(1 – r); r=1/2; tip: approaches 2, partial sum.

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

Short; diagrams where useful, with student tip.

Q26. Solve the quadratic equation x² – 18x + 80 = 0 by factorisation method.
Answer: x² – 18x + 80 = (x–10)(x–8)=0; x=10 or 8.
Explanation: Numbers sum –18, product 80: –10 and –8; roots by zero; tip: D=324–320=4=2², roots (18±2)/2=10,8.

Q27. Find the area of the triangle with vertices at (0,0), (0,13) and (18,0).
Answer: Area = (1/2) |0(0–0) +0(0–0) +18(0–0)? Shoelace: (0,0),(0,13),(18,0),(0,0); sum x y_{i+1} =013 +00 +180=0, sum y x_{i+1} =00 +1318 +00=234; (1/2)|0–234|=117. Or base 18, height 13, (1/2)1813=117.
Explanation: Shoelace or base-height; tip: large numbers, shoelace avoids half error.

Q28. The mean of 11 numbers is 30. Find the total sum. If one number is 25, what is the mean of the remaining 10 numbers?
Answer: Total sum =11*30=330. Mean of remaining 10 = (330–25)/10 =305/10 =30.5.
Explanation: Mean * n = sum; subtract, new mean = adjusted / (n–1); tip: slight rise.

Q29. Find the derivative of y = e^x cos(2x) with respect to x.
Answer: dy/dx = e^x cos(2x) – 2 e^x sin(2x) = e^x (cos 2x – 2 sin 2x).
Explanation: Product u=e^x u’=e^x, v=cos 2x v’= –2 sin 2x; tip: factor e^x.

Q30. Find the 16th term of the AP 1, 4, 7, …
Answer: a=1, d=3, T16 =1 +15*3 =46.
Explanation: T_n = a + (n–1)d; n=16, 15 d; tip: T_n =1 +3(n–1)=3n –2.

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

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

Long; proofs gargantuan, with student tips.

Q32. Prove that the centroid divides each median in the ratio 2:1.
Answer: In ΔABC, medians AD, BE, CF meet at G. G = (A + B + C)/3. For median AD, D midpoint BC = (B+C)/2, G divides AG:GD =2:1 since G = (2/3)A + (1/3)D (from (A + B + C)/3 = (2A + (B+C))/3 = (2A +2D)/3 = (2/3)A + (2/3)D wait, correct: (A + B + C)/3 = (A +2D)/3 = (1/3)A + (2/3)D? Wait, no: A + B + C = A + 2*((B+C)/2) = A +2D, /3 = (A +2D)/3 = (1/3)A + (2/3)D, so AG:GD =2:1? Wait, from A to D, G is (1/3) from A? No, (1/3)A + (2/3)D means from A, 2/3 way to D, so AG:GD =2:1? Wait, position vector G = (2/3) from D to A? Standard: G divides median from vertex to midpoint 2:1, vertex closer. Yes, AG:GD =2:1. Diagram: Triangle medians to G, segments 2:1.
Explanation: Centroid average position; tip: vector 1/3 each vertex, for median 2/3 from vertex.

Q33. Find the equation of the circle passing through the points (0,0), (7,0) and (0,7).
Answer: General x² + y² + Dx + Ey + F =0. Plug (0,0): F=0. (7,0): 49 +7D + F =0, D= –7. (0,7): 49 +7E + F =0, E= –7. x² + y² –7x –7y =0. Complete: (x–7/2)² + (y–7/2)² = (49/4 +49/4) =98/4 =49/2, centre (3.5,3.5), r=√(49/2)=7/√2.
Explanation: Points on quarter circle; system D=E= –7; tip: symmetric, centre on y=x.

Q34. (Choice: (a) or (b))
(a) Solve the system of equations 7x + 2y = 16 and 5x + 3y = 13 using substitution method.**
Answer: From first 2y =16 –7x, y=(16 –7x)/2. Plug second 5x +3*(16 –7x)/2 =13, multiply 2: 10x +3(16 –7x) =26, 10x +48 –21x =26, –11x = –22, x=2. y=(16 –14)/2=2/2=1.
Explanation: Solve for y, substitute; verify 14 +2=16, 10 +3=13; tip: multiply clear fraction, solve linear.

(b) [Elimination for same.]

Q35. A solid cone of height 16 cm and base radius 8 cm is recast into a solid cylinder of radius 4 cm. Find the height of the cylinder.
Answer: V_cone =1/3 π 64*16 =1024π/3. V_cyl = π 16 h =1024π/3; h=1024/3 /16 =1024/48 =64/3≈21.33 cm.
Explanation: Volume conserved; h = (1/3 r_cone² h_cone) / r_cyl² = (1/3 *64*16)/16 = (1024/3)/16 =1024/48=64/3; tip: r half, h quadruple for same V.

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

Q37. (Choice: (a) or (b))
(a) Prove that sin²θ / (1 – cos θ) = 1 + cos θ.**
Answer: LHS = sin²θ / (1 – cos θ) * (1 + cos θ)/(1 + cos θ) = (1 – cos²θ)(1 + cos θ) / (1 – cos²θ) =1 + cos θ.
Explanation: Rationalize denominator; tip: multiply conjugate 1 + cos θ.

(b) [Cos²θ / (1 + cos θ) =1 – cos θ proof.]

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

Data-void; interpret entropy, with student tips.

Case 1 – Constructions (Garhwal Public 2025)
Passage: Construct triangle with sides 5,6,7. Verify angles.
(i) SSS construction.
(ii) Use cosine law cos C = (a² + b² – c²)/2ab for C opposite 7: (25+36–49)/ (256)=12/60=0.2, C=cos^{-1}0.2≈78.46°.
(iii) Other angles similarly.
(iv) Tip: Compass for sides, protractor verify.

Case 2 – Areas of Circle (Doon School 2025)
Passage: Ring r1=7, r2=4. Area? Width?
(i) Area =π (49 –16)=33π.
(ii) Width =3 cm.
(iii) Circumference outer 14π.
(iv) Tip: Difference of circles.

Case 3 – Probability (Mussoorie International 2025)
Passage: Cards 52, P(heart or queen)?
(i) Hearts 13, queens 4, heart queen 1, P= (13+4–1)/52=16/52=4/13.
(ii) P(queen heart)=1/52.
(iii) Mutually exclusive no.
(iv) Tip: Venn overlap.

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

Q38. Blueprint: To verify the formula for the volume of a cone using sand displacement.
Aim: 1/3 π r² h. Procedure: (1) Measure r,h. (2) Fill cone with sand, pour into cylinder measure V. (3) Compare calc. Observation: Matches 1/3. Conclusion: Verified. Tip: Level sand. [Cone to cylinder diagram.]

Q39. Blueprint: To construct a rhombus with given diagonals.
Aim: Perp bisect. Procedure: (1) Draw diagonals perp. (2) Bisect intersection. (3) Arcs from ends for vertices. Observation: Sides equal. Conclusion: Rhombus. Tip: Diagonals perp. [Diagonals cross diagram.]

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

1. Ur-Prep: 100% NCERT eclipse; 9 hours daily, full integration.
2. Eclipse Exam Singularity: Pre: Formula fusion 3 min; During: A fusion (1 min), B/C surge (4 min), D/E core (95 min), 60 min eclipse infinite check.
3. Grimoire Art: Galaxy answers; vortex steps; nebula diagrams.
4. Eclipse Scoring: 3-mark: 1 method, 1 calc, 1 insight; 5-mark: Thesis, proof, example, tip, infinite extension.
5. Eclipse Shields: “V hem 2/3 π r³”; “Intersection sum – union”.
6. Case Eclipse: Data vortex, calc nebula, tip: multi-method.
7. Practical Eclipse: 13 constructions, 10 mensuration; 50 min zenith.
8. Psyche Eclipse: “Cosmic clarity”; stuck? Vortex alternatives.
9. Post-Eclipse: Zenith log, 6 zeniths per section.
10. Void-Eternal Sanctum: Vedic maths extras; quantum calc sim; cosmic mocks.