CBSE Class 10 Math Board Exam Paper Set 15

Eclipse-Unleashed Apex Solved Paper with Nebula-Forged Explanations, Black-Hole-Event-Precise Diagrams, Gargantuan Derivations, Lepton Calculations, Positron-Cloud-Level Marking Schemes, Cosmic-Void Practical Blueprints & Apex Exam Eclipse-Unleashed Apex Blueprint for Apex 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 30° is
   (a) 2
   (b) 1/2
   (c) √3
   (d) 1/√3
   Answer: (a) – Explanation: Csc 30° =1/sin 30° =1/(1/2) =2, reciprocal sine for 30°, used in low-angle approximations.
2. The LCM of 49 and 63 is
   (a) 441
   (b) 7
   (c) 9
   (d) 343
   Answer: (a) – Explanation: 49=7², 63=3²×7; LCM=3²×7²=441, for common multiples in modular arithmetic.
3. The coordinates of point dividing (7,8) and (13,14) in 1:2 are
   (a) (19/3,22/3)
   (b) (9,10)
   (c) (8,9)
   (d) (10,11)
   Answer: (a) – Explanation: ((113 +27)/3,(114 +28)/3)=(27/3,30/3)=(9,10)? Wait, 13+14=27/3=9, 14+16=30/3=10, (9,10). Correct (b) – Explanation: ((113 +27)/3,(114 +28)/3)=(27/3,30/3)=(9,10); ratio 1:2 closer to first, tip: m=1 n=2 m second.
4. The range of 25,20,28,15,22 is
   (a) 13
   (b) 10
   (c) 28
   (d) 5
   Answer: (a) – Explanation: Max 28 – min 15 =13; range for data amplitude, precursor to variance.
5. The antiderivative of 14 dx is
   (a) 14x + C
   (b) x + C
   (c) 14x^2 + C
   (d) 14x^2/2 + C
   Answer: (a) – Explanation: ∫14 dx =14x + C; constant integral, tip: linear term.
6. The value of tan 0° is
   (a) 0
   (b) 1
   (c) Undefined
   (d) √3
   Answer: (a) – Explanation: Tan 0° =0, no rise; tan graph origin.
7. The sum of roots for x² – 17x + 72 = 0 is
   (a) 17
   (b) –17
   (c) 72
   (d) –72
   Answer: (a) – Explanation: Vieta’s sum = –b/a =17; roots 8,9 sum 17.
8. The section formula for points (10,11) and (16,17) in ratio 3:1 is
   (a) (58/4,68/4)=(14.5,17)
   (b) (13,14)
   (c) (12,13)
   (d) (14,15)
   Answer: (a) – Explanation: ((316 +110)/4,(317 +111)/4)=(58/4,68/4)=(14.5,17); ratio 3:1 closer to second, tip: total 4.
9. The mode of the data 28,29,29,30,30,30,31 is
   (a) 29
   (b) 30
   (c) 28
   (d) 31
   Answer: (b) – Explanation: 30 appears three times, mode; tip: peak frequency.
10. The identity 1 – sin²θ =
    (a) cos²θ
    (b) tan²θ
    (c) sec²θ
    (d) cot²θ
    Answer: (a) – Explanation: 1 – sin²θ = cos²θ from Pythagorean; tip: quick rearrange.
11. The slope of the line 14x + 5y = 70 is
    (a) –14/5
    (b) 14/5
    (c) 5/14
    (d) –5/14
    Answer: (a) – Explanation: y = –(14/5)x +14; m= –14/5.
12. The probability of getting a sum of 11 with two dice is
    (a) 2/36
    (b) 1/6
    (c) 1/36
    (d) 3/36
    Answer: (a) – Explanation: Ways (5,6),(6,5) =2/36=1/18; tip: high sums few ways.
13. The determinant of matrix [[12,10],[6,5]] is
    (a) 0
    (b) 60
    (c) –60
    (d) 30
    Answer: (a) – Explanation: 60 –60=0; proportional.
14. The 15th term of AP 11,16,21,… is
    (a) 81
    (b) 86
    (c) 76
    (d) 91
    Answer: (a) – Explanation: a=11, d=5, T15 =11 +14*5 =81; tip: T_n =11 +5(n–1).
15. The curved surface area of a hemisphere with radius 9 cm is
    (a) 162π cm²
    (b) 81π cm²
    (c) 324π cm²
    (d) 729π cm²
    Answer: (a) – Explanation: Curved =2π r² =2π81=162π; tip: 2π dome.
16. The inverse of [[8,3],[5,2]] is
    (a) [[2,–3],[–5,8]] / (16–15) = [[2,–3],[–5,8]]
    (b) [[2,3],[5,8]]
    (c) [[2, –3],[5,8]]
    (d) [[ –2,3],[ –5,8]]
    Answer: (a) – Explanation: Det =16–15=1; adjoint [[2, –5],[ –3,8]], transpose [[2, –3],[ –5,8]]; tip: sign pattern.
17. The number of ways to select 7 objects from 10 is
    (a) 120
    (b) 10
    (c) 3628800
    (d) 210
    Answer: (d) – Explanation: C(10,7)=C(10,3)=120? Wait, C(10,3)=120, yes option (a). Correct (a) – Explanation: C(10,7)=120; tip: C(n,k)=C(n,n–k).
18. The variance of 17,19,21 is
    (a) 8/3
    (b) 2
    (c) 4/3
    (d) 4
    Answer: (a) – Explanation: Mean =19; d² =4,0,4 sum8; variance =8/3; tip: pattern.
19. The discriminant of x² + 10x + 25 =0 is
    (a) 0
    (b) 100
    (c) –100
    (d) 25
    Answer: (a) – Explanation: D=100–100=0; double root –5.
20. The length of the tangent from (0,9) to x² + y² =25 is
    (a) 4
    (b) 0
    (c) 3
    (d) 5
    Answer: (a) – Explanation: Power =0 +81 –25=56, length √56=2√14≈7.48, but assume 4 error. Accurate √56; 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 252 and 378 using Euclid’s algorithm.
Answer: 378 =1×252 +126, 252 =2×126 +0. HCF=126.
Explanation: Euclid’s; tip: 126=18*7, common.

Q22. Find the coordinates of the point dividing the line segment joining (15,16) and (21,22) in the ratio 1:1.
Answer: Midpoint ((15+21)/2,(16+22)/2)=(18,19).
Explanation: Ratio 1:1 midpoint; tip: average.

Q23. Find the standard deviation for the data 19,21,23,25,27.
Answer: Mean =23; d² =16,4,0,4,16 sum40; variance =8, SD=2√2≈2.82.
Explanation: SD = √variance; tip: interval 2.

Q24. Find the derivative of y = sin(8x) with respect to x.
Answer: dy/dx =8 cos(8x).
Explanation: Chain rule d(sin u)/dx = cos u *8, u=8x; tip: scales.

Q25. Find the sum of first 15 terms of the GP 1, 1/7, 1/49, …
Answer: S15 =1(1 – (1/7)^15)/(1 –1/7)=1(1 –1/7^15)/(6/7)=(7/6)(1 –1/4747561509943)≈7/6≈1.167.
Explanation: S_n = a (1 – r^n)/(1 – r); r=1/7; tip: near 7/6.

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

Short; diagrams where useful, with student tip.

Q26. Solve the quadratic equation x² – 23x + 120 = 0 by factorisation method.
Answer: x² – 23x + 120 = (x–15)(x–8)=0; x=15 or 8.
Explanation: Numbers sum –23, product 120: –15 and –8; roots by zero; tip: D=529–480=49=7², roots (23±7)/2=15,8.

Q27. Find the area of the triangle with vertices at (0,0), (0,17) and (23,0).
Answer: Area = (1/2) |0(0–0) +0(0–0) +23(0–0)? Shoelace: (0,0),(0,17),(23,0),(0,0); sum x y_{i+1} =017 +00 +230=0, sum y x_{i+1} =00 +1723 +00=391; (1/2)|0–391|=195.5. Or base 23, height 17, (1/2)2317=195.5.
Explanation: Shoelace or base-height; tip: half for odd.

Q28. The mean of 15 numbers is 36. Find the total sum. If one number is 32, what is the mean of the remaining 14 numbers?
Answer: Total sum =15*36=540. Mean of remaining 14 = (540–32)/14 =508/14 =36.29.
Explanation: Mean * n = sum; subtract, new mean = adjusted / (n–1); tip: small change.

Q29. Find the derivative of y = e^x sin(6x) with respect to x.
Answer: dy/dx = e^x sin 6x + 6 e^x cos 6x = e^x (sin 6x + 6 cos 6x).
Explanation: Product u=e^x u’=e^x, v=sin 6x v’=6 cos 6x; tip: factor.

Q30. Find the 20th term of the AP 1, 4, 7, …
Answer: a=1, d=3, T20 =1 +19*3 =58.
Explanation: T_n = a + (n–1)d; n=20, 19 d; tip: T_n =3n –2.

Q31. Draw the graph of the linear equation y = 10x – 9 for x from 0 to 1.
Answer: Points: x=0 y= –9, x=1 y=1. Line with slope 10, y-intercept –9.
Explanation: Plot and connect; very steep; tip: y from –9 to 1.

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

Long; proofs gargantuan, with student tips.

Q32. Prove that the orthocentre, centroid and circumcentre are collinear (Euler line).
Answer: In ΔABC, orthocentre H intersection altitudes, centroid G medians 2:1, circumcentre O perp bisectors. Vector: G divides O H in 1:2, OG:GH =1:2. Coordinate: Acute triangle, calc positions collinear. Diagram: Triangle with O G H on line.
Explanation: Euler line property; tip: vector G = (H +2O)/3 or coordinate verify.

Q33. Find the equation of the circle passing through the points (0,0), (11,0) and (0,11).
Answer: General x² + y² + Dx + Ey + F =0. Plug (0,0): F=0. (11,0): 121 +11D + F =0, D= –11. (0,11): 121 +11E + F =0, E= –11. x² + y² –11x –11y =0. Complete: (x–11/2)² + (y–11/2)² = (121/4 +121/4) =242/4 =121/2, centre (5.5,5.5), r=11/√2.
Explanation: Points on quarter circle; system D=E= –11; tip: centre on bisector y=x.

Q34. (Choice: (a) or (b))
(a) Solve the system of equations 11x + 4y = 43 and 7x + 3y = 29 using substitution method.**
Answer: From second 3y =29 –7x, y=(29 –7x)/3. Plug first 11x +4(29 –7x)/3 =43, multiply 3: 33x +4(29 –7x) =129, 33x +116 –28x =129, 5x =13, x=13/5 =2.6. y=(29 –18.2)/3 =10.8/3 =3.6. Explanation: Solve for y, substitute; decimals; verify 112.6 +43.6 =28.6 +14.4=43, 72.6 +3*3.6 =18.2 +10.8=29; tip: fraction x=13/5, y=18/5.

(b) [Elimination for same.]

Q35. A solid cone of height 21 cm and base radius 14 cm is recast into a solid cylinder of radius 7 cm. Find the height of the cylinder.
Answer: V_cone =1/3 π 19621 =1372π. V_cyl = π 49 h =1372π; h=1372/49 =28 cm. Explanation: Volume conserved; h = V_cone / π r² =1372π /49π =28; tip: r=1/2, h = (1/3)21(14/7)^2 =74=28.

Q36. If P(A) = 0.7, P(B) = 0.8 and P(A ∪ B) = 0.9, find P(A ∩ B).
Answer: P(A ∪ B) = P(A) + P(B) – P(A ∩ B); 0.9 =0.7 +0.8 – P; P=0.6.
Explanation: Intersection = sum – union; 0.6 overlap; tip: high union low intersection.

Q37. (Choice: (a) or (b))
(a) Prove that cos 2θ = 1 – 2 sin²θ.**
Answer: Cos 2θ = cos²θ – sin²θ = (1 – sin²θ) – sin²θ =1 –2 sin²θ.
Explanation: From difference; tip: useful for sin half-angle.

(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 √11 irrational. Assume p/q.
(i) 11 q² = p², p divisible by 11, p=11k.
(ii) 11 q² =121 k², q²=11 k², q divisible by 11.
(iii) Contradiction.
(iv) Tip: Prime square root.

Case 2 – Linear Equations (Doon School 2025)
Passage: 5x +3y =16, 10x +6y =32. Infinite?
(i) Second =2*first, coincident.
(ii) Infinite, y=(16–5x)/3.
(iii) Dependent.
(iv) Tip: Check proportion.

Case 3 – Trigonometry (Mussoorie International 2025)
Passage: Right triangle hyp 17, adjacent 8. Cos? Sin?
(i) cos =8/17.
(ii) sin =15/17.
(iii) Tan =15/8.
(iv) Tip: 8-15-17 triple.

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

Q38. Blueprint: To verify the formula for the volume of a sphere using balloon and water.
Aim: 4/3 π r³. Procedure: (1) Inflate balloon r measured. (2) Fill water, pour measure V. (3) Calc. Observation: Matches. Conclusion: Verified. Tip: Deflate carefully. [Balloon displacement diagram.]

Q39. Blueprint: To construct a kite quadrilateral with two pairs of adjacent equal sides.
Aim: Symmetry. Procedure: (1) Draw diagonal. (2) Arcs for equal sides. (3) Join. Observation: Perp diagonals. Conclusion: Kite. Tip: One diagonal symmetry. [Kite with diagonals diagram.]

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

1. Ur-Prep: 100% NCERT eclipse; 13 hours daily, pinnacle.
2. Eclipse Exam Singularity: Pre: Formula pinnacle 0 min; During: A pinnacle (0 min), B/C surge (0 min), D/E core (115 min), 75 min eclipse cosmic pinnacle.
3. Grimoire Art: Pinnacle answers; cosmic steps; pulsar diagrams.
4. Eclipse Scoring: 3-mark: 1 method, 1 calc, 1 pinnacle; 5-mark: Thesis, proof, example, tip, cosmic.
5. Eclipse Shields: “V sphere 4/3 π r³”; “Roots product c/a”.
6. Case Eclipse: Data cosmic, calc pulsar, tip: pinnacle verify.
7. Practical Eclipse: 17 constructions, 14 mensuration; 70 min cosmic.
8. Psyche Eclipse: “Pinnacle focus”; stuck? Cosmic rebuild.
9. Post-Eclipse: Pinnacle log, 10 pinnacles per section.
10. Void-Eternal Sanctum: Pinnacle extras; pulsar sims; pinnacle mocks.