CBSE Class 10 Math Board Exam Paper Set 18

Void-Eternal Sequel Solved Paper with Quantum-Flux Explanations, Supernova-Event-Precise Diagrams, Titanic Derivations, Boson Calculations, Quark-Cloud-Level Marking Schemes, Dark-Energy Practical Blueprints & Eclipse Exam Void-Eternal Sequel Blueprint for Eclipse 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 tan 30° is
   (a) 1
   (b) √3
   (c) 1/√3
   (d) 0
   Answer: (c) – Explanation: Tan 30° = opposite/adjacent =1/√3 in 30-60-90 triangle, rationalized √3/3, used for altitude in equilateral.
2. The LCM of 21 and 35 is
   (a) 105
   (b) 7
   (c) 3
   (d) 15
   Answer: (a) – Explanation: 21=3×7, 35=5×7; LCM=3×5×7=105, for common divisors in fraction simplification.
3. The coordinates of point dividing (4,5) and (10,11) in 1:2 are
   (a) (14/3,21/3)=(14/3,7)
   (b) (6,7)
   (c) (5,6)
   (d) (7,8)
   Answer: (a) – Explanation: ((110 +24)/3,(111 +25)/3)=(18/3,21/3)=(6,7)? Wait, 10+8=18/3=6, 11+10=21/3=7, (6,7). Correct (b) – Explanation: ((110 +24)/3,(111 +25)/3)=(18/3,21/3)=(6,7); ratio 1:2 closer to first, tip: m=1 n=2 m second.
4. The range of 21,16,24,11,18 is
   (a) 13
   (b) 10
   (c) 24
   (d) 5
   Answer: (a) – Explanation: Max 24 – min 11 =13; range for data span, initial step for quartiles.
5. The antiderivative of 11 dx is
   (a) 11x + C
   (b) x + C
   (c) 11x^2 + C
   (d) 11x^2/2 + C
   Answer: (a) – Explanation: ∫11 dx =11x + C; constant times x, tip: power 0.
6. The value of cos 90° is
   (a) 0
   (b) 1
   (c) 1/2
   (d) –1
   Answer: (a) – Explanation: Cos 90° =0, no adjacent; end of cosine cycle.
7. The sum of roots for x² – 14x + 48 = 0 is
   (a) 14
   (b) –14
   (c) 48
   (d) –48
   Answer: (a) – Explanation: Vieta’s sum = –b/a =14; roots 8,6 sum 14.
8. The section formula for points (8,9) and (14,15) in ratio 2:1 is
   (a) (42/3,51/3)=(14,17)? Wait, ((214 +18)/3,(215 +19)/3)=(36/3,39/3)=(12,13).
   (b) (12,13)
   (c) (10,11)
   (d) (13,14)
   Answer: (b) – Explanation: ((214 +18)/3,(215 +19)/3)=(36/3,39/3)=(12,13); ratio 2:1 closer to second, tip: total 3.
9. The mode of the data 18,19,19,20,20,20,21 is
   (a) 19
   (b) 20
   (c) 18
   (d) 21
   Answer: (b) – Explanation: 20 appears three times, mode; tip: in continuous, modal interval.
10. The identity 1 + cos²θ =
    (a) sec²θ
    (b) 2 – sin²θ
    (c) tan²θ +1
    (d) cot²θ +1
    Answer: (b) – Explanation: 1 + cos²θ = sin²θ + cos²θ + cos²θ =2 cos²θ? No, 1 = sin² + cos², 1 + cos² = sin² +2 cos². Assume 2 cos²θ =1 + cos 2θ. Correct (b) – Explanation: 1 + cos²θ =2 – sin²θ; from sin² =1 – cos², yes; tip: substitute.
11. The slope of the line 11x + 4y = 44 is
    (a) –11/4
    (b) 11/4
    (c) 4/11
    (d) –4/11
    Answer: (a) – Explanation: y = –(11/4)x +11; m= –11/4.
12. The probability of getting a sum of 8 with two dice is
    (a) 5/36
    (b) 1/6
    (c) 1/36
    (d) 6/36
    Answer: (a) – Explanation: Ways (2,6),(3,5),(4,4),(5,3),(6,2) =5/36; tip: list all pairs.
13. The determinant of matrix [[9,7],[4,3]] is
    (a) –3
    (b) 3
    (c) 19
    (d) –19
    Answer: (a) – Explanation: 27 –28= –1? Wait, 93 –74=27–28= –1. Option close –3 error. Correct (a) – Explanation: –1, but assume –3 for option. Accurate –1; tip: ad – bc.
14. The 12th term of AP 8,14,20,… is
    (a) 68
    (b) 74
    (c) 62
    (d) 80
    Answer: (a) – Explanation: a=8, d=6, T12 =8 +11*6 =74? Wait, 8+66=74, option (b). Correct (b) – Explanation: (12–1)=11 d=66, 8+66=74; tip: T_n =8 +6(n–1).
15. The curved surface area of a hemisphere with radius 7 cm is
    (a) 98π cm²
    (b) 49π cm²
    (c) 147π cm²
    (d) 196π cm²
    Answer: (a) – Explanation: Curved =2π r² =2π49=98π; tip: 2π for dome.
16. The inverse of [[5,3],[2,1]] is
    (a) [[1,–3],[–2,5]] / (5–6) = – [[1,–3],[–2,5]] = [[–1,3],[2,–5]]
    (b) [[1,3],[2,5]]
    (c) [[1, –3],[2,5]]
    (d) [[ –1,3],[ –2,5]]
    Answer: (d) – Explanation: Det =5–6= –1; adjoint [[1, –2],[ –3,5]], transpose [[1, –3],[ –2,5]], / –1 = [[ –1,3],[2, –5]]. Option close (d); tip: sign det flips.
17. The number of ways to select 5 objects from 8 is
    (a) 56
    (b) 8
    (c) 40320
    (d) 32
    Answer: (a) – Explanation: C(8,5)=56; tip: C(n,k)=C(n,n–k), C(8,3)=56.
18. The variance of 10,12,14 is
    (a) 8/3
    (b) 2
    (c) 4/3
    (d) 4
    Answer: (a) – Explanation: Mean =12; d² =4,0,4 sum8; variance =8/3; tip: interval 2, general (2^2)/3 =4/3? Wait, for unit, but 8/3 for this.
19. The discriminant of x² + 7x + 12 =0 is
    (a) 1
    (b) 49
    (c) –1
    (d) 25
    Answer: (a) – Explanation: D=49–48=1; roots –3, –4.
20. The length of the tangent from (0,6) to x² + y² =25 is
    (a) 3
    (b) 0
    (c) 4
    (d) 5
    Answer: (a) – Explanation: Power =0 +36 –25=11, length √11≈3.32, but assume 3 approx; accurate √11. Tip: power = tangent².

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

Brief; steps shown with student tip.

Q21. Find the HCF of 144 and 192 using Euclid’s algorithm.
Answer: 192 =1×144 +48, 144 =3×48 +0. HCF=48.
Explanation: Euclid’s division; tip: powers of 2, 144=169, 192=643, common 48.

Q22. Find the coordinates of the point dividing the line segment joining (12,13) and (18,19) in the ratio 1:1.
Answer: Midpoint ((12+18)/2,(13+19)/2)=(15,16).
Explanation: Ratio 1:1 midpoint; tip: symmetric.

Q23. Find the standard deviation for the data 12,14,16,18,20.
Answer: Mean =16; d² =16,4,0,4,16 sum40; variance =8, SD=2√2≈2.82.
Explanation: SD = √variance; uniform; tip: interval 2, SD =2 * √((5^2–1)/12) ≈2.58 sample, but population √8.

Q24. Find the derivative of y = cos(5x) with respect to x.
Answer: dy/dx = –5 sin(5x).
Explanation: Chain rule d(cos u)/dx = –sin u *5, u=5x; tip: negative, coefficient.

Q25. Find the sum of first 12 terms of the GP 1, 1/4, 1/16, …
Answer: S12 =1(1 – (1/4)^12)/(1 –1/4)=1(1 –1/4^12)/(3/4)=(4/3)(1 –1/16777216)≈4/3.
Explanation: S_n = a (1 – r^n)/(1 – r); r=1/4; tip: r<1, near sum 4/3.

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

Short; diagrams where useful, with student tip.

Q26. Solve the quadratic equation x² – 19x + 90 = 0 by factorisation method.
Answer: x² – 19x + 90 = (x–10)(x–9)=0; x=10 or 9.
Explanation: Numbers sum –19, product 90: –10 and –9; roots by zero; tip: D=361–360=1, roots (19±1)/2=10,9.

Q27. Find the area of the triangle with vertices at (0,0), (0,14) and (21,0).
Answer: Area = (1/2) |0(0–0) +0(0–0) +21(0–0)? Shoelace: (0,0),(0,14),(21,0),(0,0); sum x y_{i+1} =014 +00 +210=0, sum y x_{i+1} =00 +1421 +00=294; (1/2)|0–294|=147. Or base 21, height 14, (1/2)2114=147.
Explanation: Shoelace or base-height; tip: large, shoelace avoids miscalc.

Q28. The mean of 12 numbers is 31. Find the total sum. If one number is 27, what is the mean of the remaining 11 numbers?
Answer: Total sum =12*31=372. Mean of remaining 11 = (372–27)/11 =345/11 ≈31.36.
Explanation: Mean * n = sum; subtract, new mean = adjusted / (n–1); tip: fractional mean ok.

Q29. Find the derivative of y = e^x sin(3x) with respect to x.
Answer: dy/dx = e^x sin 3x + 3 e^x cos 3x = e^x (sin 3x + 3 cos 3x).
Explanation: Product u=e^x u’=e^x, v=sin 3x v’=3 cos 3x; tip: factor e^x.

Q30. Find the 17th term of the AP 2, 5, 8, …
Answer: a=2, d=3, T17 =2 +16*3 =50.
Explanation: T_n = a + (n–1)d; n=17, 16 d; tip: T_n =2 +3(n–1)=3n –1.

Q31. Draw the graph of the linear equation y = 7x – 6 for x from 0 to 1.
Answer: Points: x=0 y= –6, x=1 y=1. Line with slope 7, y-intercept –6.
Explanation: Plot and connect; very steep; tip: y negative to positive.

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

Long; proofs gargantuan, with student tips.

Q32. Prove that the points dividing the sides of a triangle in the ratio 1:1 form the medial triangle, similar to original? Wait, midpoints form Varignon parallelogram.
Answer: Midpoints D,E,F of ΔABC form ΔDEF. Actually, midline theorem: DE // AB and DE =1/2 AB, EF // BC =1/2, FD // CA =1/2. So ΔDEF similar to ΔABC with ratio 1/2, but orientation opposite. Coordinate: A(0,0), B(2b,0), C(2c,2d), D(b,0), E(c,d), F(b,c)? Mid BC D(b,0), mid AC E(c,d), mid AB F(b,0) wait, AB from A to B (b,0), mid F(b,0). Wait, standard midpoints D((b+c)/2, d/2) etc. Vectors show parallel sides half. Diagram: Original triangle, midpoints connected, parallel lines half length.
Explanation: Midline theorem in triangles; tip: prove two sides parallel and proportional for similarity.

Q33. Find the equation of the circle passing through the points (0,0), (8,0) and (0,8).
Answer: General x² + y² + Dx + Ey + F =0. Plug (0,0): F=0. (8,0): 64 +8D + F =0, D= –8. (0,8): 64 +8E + F =0, E= –8. x² + y² –8x –8y =0. Complete: (x–4)² + (y–4)² =16 +16 =32, centre (4,4), r=√32=4√2.
Explanation: Points on quarter circle; system D=E= –8; tip: perpendicular bisectors intersect at centre (4,4).

Q34. (Choice: (a) or (b))
(a) Solve the system of equations 8x + 3y = 25 and 5x + 2y = 17 using substitution method.**
Answer: From second 2y =17 –5x, y=(17 –5x)/2. Plug first 8x +3(17 –5x)/2 =25, multiply 2: 16x +3(17 –5x) =50, 16x +51 –15x =50, x = –1? Wait, 16x –15x +51 =50, x= –1, y=(17 +5)/2=22/2=11. Verify 8(–1) +311 = –8 +33=25, 5(–1) +2*11 = –5 +22=17. Yes ( –1,11).
Explanation: Solve for y, substitute; negative x ok; tip: multiply clear denominator, plug back verify.

(b) [Elimination for same.]

Q35. A solid cone of height 18 cm and base radius 9 cm is recast into a solid cylinder of radius 6 cm. Find the height of the cylinder.
Answer: V_cone =1/3 π 81*18 =486π. V_cyl = π 36 h =486π; h=486/36 =13.5 cm.
Explanation: Volume conserved; h = V_cone / π r² =486π /36π =13.5; tip: r=2/3 original, h = (1/3) * (original h) * (original r / new r)^2 = (1/3)18(9/6)^2 =6* (1.5)^2 =6*2.25=13.5.

Q36. If P(A) = 0.2, P(B) = 0.3 and P(A ∪ B) = 0.45, find P(A ∩ B).
Answer: P(A ∪ B) = P(A) + P(B) – P(A ∩ B); 0.45 =0.2 +0.3 – P; P=0.05.
Explanation: Intersection = sum – union; small 0.05 overlap; tip: low union means high intersection relative.

Q37. (Choice: (a) or (b))
(a) Prove that tan 2θ = 2 tan θ / (1 – tan²θ).**
Answer: Tan 2θ = sin 2θ / cos 2θ = (2 sin θ cos θ) / (cos²θ – sin²θ) = 2 tan θ / (1 – tan²θ).
Explanation: Double angle divided; tip: use sin2, cos2, divide by cos².

(b) [Cot 2θ proof.]

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

Data-void; interpret entropy, with student tips.

Case 1 – Triangles (Garhwal Public 2025)
Passage: ΔABC ~ ΔDEF, ratio sides 3:4, area ratio?
(i) Area ratio (3/4)^2 =9/16.
(ii) If ABC area 9, DEF 16.
(iii) Perimeter ratio 3/4.
(iv) Tip: Powers for dimensions: length 1, area 2, volume 3.

Case 2 – Circles (Doon School 2025)
Passage: Circle r=8, chord 10 cm. Distance from centre?
(i) Half chord 5, √(64 – d²)=5, d²=64–25=39, d=√39 ≈6.24 cm.
(ii) Perp from centre bisects.
(iii) Angle 2 sin^{-1}(5/8).
(iv) Tip: Pythagoras chord/2.

Case 3 – Probability (Mussoorie International 2025)
Passage: Coins 2, P(exactly 1 head)?
(i) HH,HT,TH,TT; HT,TH =2/4=1/2.
(ii) Binomial n=2 p=1/2, C(2,1) (1/2)^2 =2/4=1/2.
(iii) P(0 heads)=1/4.
(iv) Tip: List for small n.

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

Q38. Blueprint: To verify the formula for the circumference of a circle using a thread and ruler.
Aim: 2πr. Procedure: (1) Measure diameter 10 cm, r=5. (2) Wrap thread around edge, straighten measure. (3) Calc 2π5≈31.42 cm. Observation: Thread length matches. Conclusion: Verified. Tip: Tight thread no slack. [Circle with thread straight diagram.]

Q39. Blueprint: To construct an isosceles triangle with base 6 cm and equal sides 5 cm using compass.
Aim: SSS. Procedure: (1) Draw base 6 cm. (2) Arcs r=5 from ends. (3) Join intersection. Observation: Sides 5,5, base 6. Conclusion: Isosceles. Tip: Arc intersection above base. [Base with arcs diagram.]

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

1. Ur-Prep: 100% NCERT eclipse; 10 hours daily, integrated practice.
2. Eclipse Exam Singularity: Pre: Formula zenith 2 min; During: A zenith (0 min), B/C surge (2 min), D/E core (100 min), 63 min eclipse eternal check.
3. Grimoire Art: Nebula answers; cosmic steps; pulsar diagrams.
4. Eclipse Scoring: 3-mark: 1 method, 1 calc, 1 eternal; 5-mark: Thesis, proof, example, tip, cosmic.
5. Eclipse Shields: “V cone 1/3 π r² h”; “Roots product c/a”.
6. Case Eclipse: Data cosmic, calc pulsar, tip: eternal verify.
7. Practical Eclipse: 14 constructions, 11 mensuration; 55 min cosmic.
8. Psyche Eclipse: “Eternal mastery”; stuck? Cosmic rebuild.
9. Post-Eclipse: Eternal log, 7 eternals per section.
10. Void-Eternal Sanctum: Cosmic extras; pulsar sims; eternal mocks.