CBSE Class 10 Math Board Exam Paper Set 6

Zenith-Unleashed Solved Paper with Cosmic-Nexus Explanations, Pulsar-Event-Precise Diagrams, Titanic Derivations, Hadron Calculations, Electron-Nexus-Level Marking Schemes, Nebula-Vortex Practical Blueprints & Nova Exam Zenith-Unleashed Blueprint for Nova 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 cos 45° is
   (a) 1/2
   (b) √2/2
   (c) √3/2
   (d) 0
   Answer: (b) – Explanation: Cos 45° = adjacent/hypotenuse in isosceles right triangle = √2/2, essential for diagonal calculations in squares and 45° rotations.
2. The LCM of 8 and 12 is
   (a) 24
   (b) 4
   (c) 48
   (d) 96
   Answer: (a) – Explanation: 8=2^3, 12=2^2×3; LCM=2^3×3=24, key for common multiples in time cycles or fraction denominators.
3. The coordinates of point dividing (0,0) and (8,12) in 2:1 are
   (a) (16/3,8)
   (b) (16/3,8) wait, ((28 +10)/3,(212 +10)/3)=(16/3,24/3)=(16/3,8).
   (c) (5.33,8)
   (d) (8,5.33)
   Answer: (a) – Explanation: Section formula ((28 +10)/3,(212 +10)/3)=(16/3,8); weighted average, closer to second point, tip: multiply by ratio parts.
4. The range of 11,6,14,3,9 is
   (a) 11
   (b) 8
   (c) 14
   (d) 6
   Answer: (a) – Explanation: Max 14 – min 3 =11; basic spread measure, helps identify data variability like scores in a class.
5. The antiderivative of x^2 dx is
   (a) x^3/3 + C
   (b) x^3 + C
   (c) 3x^3 + C
   (d) x^3/3
   Answer: (a) – Explanation: ∫x^2 dx = x^3/3 + C; power rule (n+1 in denominator), tip: forget /3 and lose marks, always include C.
6. The value of sin 60° is
   (a) 1/2
   (b) √3/2
   (c) √2/2
   (d) 1
   Answer: (b) – Explanation: Sin 60° = opposite/hypotenuse in equilateral-derived triangle = √3/2, used in 60° angles for regular hexagons.
7. The sum of roots for x² – 5x + 6 = 0 is
   (a) 5
   (b) –5
   (c) 6
   (d) –6
   Answer: (a) – Explanation: Vieta’s sum = –b/a =5; roots 2,3 sum 5, quick tool for quadratic without full solve.
8. The section formula for points (3,4) and (9,10) in ratio 1:2 is
   (a) (5,6)
   (b) (6,7)
   (c) (7,8)
   (d) (4,5)
   Answer: (a) – Explanation: ((19 +23)/3,(110 +24)/3)=(15/3,18/3)=(5,6); ratio 1:2 closer to first, tip: n parts first if from second? No, m:n m second n first.
9. The mode of the data 1,2,2,3,3,3,4 is
   (a) 2
   (b) 3
   (c) 1
   (d) 4
   Answer: (b) – Explanation: 3 appears three times, mode most common; for bimodal, list both, tip: frequency table first for large data.
10. The identity cos²θ + sin²θ =
    (a) 1
    (b) 0
    (c) sec²θ
    (d) tan²θ
    Answer: (a) – Explanation: From unit circle radius 1, cos= x/r, sin=y/r, x² + y² = r² =1; cornerstone for trig proofs.
11. The slope of the line 2x + 3y = 6 is
    (a) –2/3
    (b) 2/3
    (c) 3/2
    (d) –3/2
    Answer: (a) – Explanation: y = –(2/3)x +2; m= –2/3, negative slope line.
12. The probability of getting an even number when a die is rolled is
    (a) 1/2
    (b) 1/3
    (c) 1/6
    (d) 2/3
    Answer: (a) – Explanation: Even 2,4,6 out of 6 faces =3/6=1/2; fair die uniform, tip: count favorable vs total.
13. The determinant of matrix [[4,5],[2,3]] is
    (a) –2
    (b) 2
    (c) –7
    (d) 7
    Answer: (a) – Explanation: ad – bc =12 –10=2? Wait, 43 –52=12–10=2, option (b). Correct (b) – Explanation: Positive 2; det for area scaling.
14. The 7th term of AP 1,4,7,… is
    (a) 19
    (b) 22
    (c) 16
    (d) 25
    Answer: (a) – Explanation: a=1, d=3, T7 =1 +6*3 =19; sequential addition, tip: pattern 1+3(k–1).
15. The volume of a sphere with radius 3 cm is
    (a) (4/3)π27
    (b) 36π
    (c) 27π
    (d) (4/3)π
    Answer: (a) – Explanation: V =4/3 π r³ =4/3 π 27 =36π; exact form preferred, tip: remember 4/3 factor.
16. The inverse of [[3,0],[0,4]] is
    (a) [[1/3,0],[0,1/4]]
    (b) [[0,3],[4,0]]
    (c) [[4,0],[0,3]]
    (d) [[1/3,4],[0,1/4]]
    Answer: (a) – Explanation: Diagonal det=12, inverse diagonals reciprocal; tip: diagonal matrices easy, multiply check I.
17. The number of ways to arrange 3 distinct books on a shelf is
    (a) 6
    (b) 3
    (c) 1
    (d) 9
    Answer: (a) – Explanation: P(3,3)=3!=6; permutations for order matters, tip: n! for all distinct.
18. The variance of 1,3,5,7 is
    (a) 5
    (b) 4
    (c) 3
    (d) 2
    Answer: (a) – Explanation: Mean=4; d²=9,1,1,9 sum20; variance =20/4=5; tip: n=4, population divide by n.
19. The discriminant of x² – 2x + 1 =0 is
    (a) 0
    (b) 4
    (c) –4
    (d) 2
    Answer: (a) – Explanation: D=4–4=0; double root 1, (x–1)².
20. The length of the tangent from the origin to the circle x² + y² =16 is
    (a) 4
    (b) 0
    (c) 16
    (d) 2
    Answer: (a) – Explanation: Power =0² +0² –16 = –16? Wait, for external point (0,0) to r=4, d=0 <r, inside, no real tangent. Correct for external point say (5,0), √(25–16)=3, but for origin inside, length 0. Option (b) 0 – Explanation: Point inside circle, no tangent from inside; tip: power d² – r² <0 no real tangent.

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

Brief; steps shown with student tip.

Q21. Find the HCF of 75 and 90 using Euclid’s algorithm.
Answer: 90 =1×75 +15, 75 =5×15 +0. HCF=15.
Explanation: Euclid’s algorithm uses repeated division; remainder becomes divisor; tip: efficient for big numbers, always larger first to minimize steps.

Q22. Find the coordinates of the point dividing the line segment joining (5,6) and (11,12) in the ratio 1:3.
Answer: ((111 +35)/4,(112 +36)/4)=(26/4,30/4)=(6.5,7.5).
Explanation: Section formula (m x2 + n x1)/(m+n); m=1 n=3 from (5,6), weighted towards first; tip: ratio 1:3 means 1 part second, 3 parts first, total 4.

Q23. Find the standard deviation for the data 4,6,8,10,12.
Answer: Mean =8; d² =16,4,0,4,16 sum40; variance =40/5=8, SD=√8=2√2≈2.82.
Explanation: SD = √[average of squared deviations]; symmetric data, easy calc; tip: for AP, SD = (n²–1)d/ (2√3 n) approximate, but direct for small n.

Q24. Find the derivative of y = cos(x^2) with respect to x.
Answer: dy/dx = –sin(x^2) * 2x.
Explanation: Chain rule d(cos u)/dx = –sin u * du/dx, u=x^2 du=2x; tip: inside derivative multiplied, common in composite functions like waves.

Q25. Find the sum of first 6 terms of the GP 1, –3, 9, –27,…
Answer: S6 =1( (–3)^6 –1 ) / (–3 –1 ) =1(729 –1)/ (–4 ) =728/ –4 = –182.
Explanation: S_n = a (r^n –1)/(r–1); r= –3; tip: for negative r, alternates, formula works, verify partial sums 1–3+9–27+81–243 = (1–3)+(9–27)+(81–243)= –2 –18 –162 = –182.

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

Short; diagrams where useful, with student tip.

Q26. Solve the quadratic equation x² – 11x + 28 = 0 by factorisation method.
Answer: x² – 11x + 28 = (x–4)(x–7)=0; x=4 or 7.
Explanation: Numbers sum –11, product 28: –4 and –7; roots by zero product; tip: factorisation for integer roots, check D=121–112=9>0, roots (11±3)/2=7,4.

Q27. Find the area of the triangle with vertices at (0,0), (0,7) and (12,0).
Answer: Area = (1/2) |0(0–0) +0(0–0) +12(0–0)? Wait, shoelace: List (0,0),(0,7),(12,0),(0,0); sum x y_{i+1} =07 +00 +120 =0, sum y x_{i+1} =00 +712 +00 =84; (1/2)|0 –84|=42. Or base 12, height 7, (1/2)127=42.
Explanation: Shoelace for coordinates or base-height; tip: order counter-clockwise for positive, absolute value.

Q28. The mean of 5 numbers is 32. Find the total sum. If one number is 28, what is the mean of the remaining 4 numbers?
Answer: Total sum =5*32=160. Mean of remaining 4 = (160–28)/4 =132/4 =33.
Explanation: Mean * n = sum; subtract, new mean = adjusted / (n–1); tip: rising mean after removing lower value, common in data adjustment.

Q29. Find the derivative of y = ln( x +1 ) with respect to x.
Answer: dy/dx = 1/(x+1).
Explanation: d(ln u)/dx = (1/u) du/dx, u=x+1 du=1; tip: logarithmic derivative simplifies products to sums, useful in growth rates.

Q30. Find the 10th term of the AP 2, 5, 8, …
Answer: a=2, d=3, T10 =2 +9*3 =2+27=29.
Explanation: T_n = a + (n–1)d; n=10, 9 intervals; tip: visualize as adding d nine times from first, for large n use formula.

Q31. Draw the graph of the linear equation y = 3x – 2 for x from 0 to 2.
Answer: Points: x=0 y= –2, x=1 y=1, x=2 y=4. Line with slope 3, y-intercept –2.
Explanation: Plot and join; steep positive slope; tip: choose easy x-values, label axes, extend for full view.

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

Long; proofs gargantuan, with student tips.

Q32. Prove that the ratio of the areas of two similar triangles is the square of the ratio of their corresponding sides.
Answer: Similar ΔABC ~ ΔDEF, AB/DE = BC/EF = CA/FD = k. Area = (1/2) base height. Height ratio k (parallel), base k, area k * k = k². Or coordinate: Place A at (0,0), B (b,0), C (p,q); area (1/2)b q. Similar scaled by k, area (1/2)(k b)(k q)=k² (1/2 b q). Diagram: Two similar triangles with corresponding sides marked k.
Explanation: Similarity preserves angles, scales sides; area scales square; tip: use vectors or coordinates for rigor, apply to circles too (π r²).

Q33. Find the equation of the circle passing through the points (2,0), (0,2) and (–2,0).
Answer: General x² + y² + Dx + Ey + F =0. Plug (2,0): 4 +2D + F =0. (0,2): 4 +2E + F =0. (–2,0): 4 –2D + F =0. Add first and third: 8 +2F =0, F= –4. From first 4 +2D –4 =0, 2D=0, D=0. From second 4 +2E –4 =0, 2E=0, E=0. x² + y² =4. Centre (0,0), r=2.
Explanation: Symmetric points on circle x² + y² =4; system solves to D=E=0; tip: plot points to guess centre, verify by distance.

Q34. (Choice: (a) or (b))
(a) Solve the system of equations 4x + 3y = 10 and 2x + y = 4 using elimination method.**
Answer: Multiply second by 3: 6x +3y =12. Subtract first: (6x +3y) – (4x +3y) =12 –10, 2x =2, x=1. From second y=4–2=2.
Explanation: Elimination cancels y; multiply to match coefficients; verify 41 +32=10, 2*1 +2=4; tip: choose variable with smaller coefficient to multiply.

(b) [Substitution for same.]

Q35. A solid cone of height 10 cm and base radius 4 cm is recast into a solid cylinder of the same base radius. Find the height of the cylinder.
Answer: V_cone =1/3 π (16)10 ≈52.33π. V_cyl = π (16) h ≈52.33π; h≈52.33/16≈3.27 cm. Accurate: 1/3 π 1610 =160π/3, h=(160π/3)/ (16π) =160/(3*16)=160/48=10/3≈3.33 cm.
Explanation: Volume conserved; h_cyl = h_cone /3 =10/3 cm; tip: same r, h scales by 1/3 for cone to cyl, easy when base fixed.

Q36. If P(A) = 0.4, P(B) = 0.3 and P(A ∪ B) = 0.6, find P(A ∩ B).
Answer: P(A ∪ B) = P(A) + P(B) – P(A ∩ B); 0.6 =0.4 +0.3 – P; P=0.1.
Explanation: Intersection from union formula; overlap 0.1; tip: if no overlap 0.7>0.6 impossible, so dependent.

Q37. (Choice: (a) or (b))
(a) Prove that cot(A – B) = (cot A cot B +1)/(cot B – cot A).**
Answer: Cot(A – B) =1/tan(A – B) = (1 – tan A tan B)/(tan A – tan B) * (cot A cot B) = (cot A cot B +1)/(cot B – cot A).
Explanation: From tan difference, reciprocate and multiply; tip: use tan form first, then convert, standard for compound angles.

(b) [Tan(A – B) proof.]

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

Data-void; interpret entropy, with student tips.

Case 1 – Polynomials (Garhwal Public 2025)
Passage: p(x)=x³ –4x² +5x –2. Possible roots ±1,2. p(1)=1–4+5–2=0. Root 1.
(i) Factor (x–1).
(ii) Divide: x² –3x +2 = (x–1)(x–2).
(iii) Roots 1 (double), 2.
(iv) Tip: Rational root theorem tests possible, synthetic division fast.

Case 2 – Lines and Angles (Doon School 2025)
Passage: Parallel lines cut by transversal, alternate interior angles equal. Prove.
(i) Co-interior sum 180°.
(ii) Alternate equal by corresponding.
(iii) Z-angles equal.
(iv) Tip: Draw transversal, label angles, use axiom.

Case 3 – Statistics (Mussoorie International 2025)
Passage: Data 15,25,35,45,55.
(i) Mean =35.
(ii) Median =35.
(iii) Mode none.
(iv) SD = √[(400+100+0+100+400)/5]=√200=10√2≈14.14. Tip: Even spacing, SD = (range/2√3) approx.

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

Q38. Blueprint: To verify the formula for the sum of an AP using graph paper.
Aim: S_n = n/2 (a + l). Procedure: (1) Plot terms vs n. (2) Cumulative sum. (3) Check linear. Observation: Matches. Conclusion: Verified. Tip: Visual for understanding. [Graph terms and sum.]

Q39. Blueprint: To construct a quadrilateral with given diagonals and angle between them.
Aim: SAS type. Procedure: (1) Draw diagonal. (2) Angle at intersection. (3) Arcs for sides. Observation: Forms quad. Conclusion: Possible. Tip: Compass precision. [Quad with diagonals diagram.]

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

1. Ur-Prep: 100% NCERT eclipse; 3 hours daily, revise trig daily, stats weekly.
2. Eclipse Exam Singularity: Pre: Formula chant 10 min; During: A blitz (7 min), B/C momentum (18 min), D/E depth (65 min), 40 min eclipse verify all.
3. Grimoire Art: Box roots; arrow steps; legend diagrams.
4. Eclipse Scoring: 3-mark: 1 method, 1 step, 1 check; 5-mark: Aim, steps, concl, tip, verify.
5. Eclipse Shields: “Shoelace absolute”; “Derivative product u’v + uv'”.
6. Case Eclipse: Data table first, compute parallel.
7. Practical Eclipse: 7 constructions, 3 mensuration; time 20 min practice.
8. Psyche Eclipse: “Precision is power”; stuck? Breathe, re-read.
9. Post-Eclipse: Error log, 1 fix per topic next session.
10. Void-Eternal Sanctum: Arihant for variants; Desmos for plots; peer review mocks.