CBSE Class 10 Math Board Exam Paper Set 8

Eclipse-Unleashed 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 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 sin 45° is
   (a) 1/2
   (b) √2/2
   (c) √3/2
   (d) 1
   Answer: (b) – Explanation: In a 45-45-90 triangle, opposite over hypotenuse is √2/2, a fundamental value for 45° angles in trigonometry, useful in coordinate geometry for lines at 45°.
2. The LCM of 12 and 18 is
   (a) 36
   (b) 6
   (c) 24
   (d) 72
   Answer: (a) – Explanation: Prime factorization 12=2²×3, 18=2×3²; LCM takes highest powers 2²×3²=36, helping in simplifying fractions or finding common multiples in real-life scheduling.
3. The coordinates of point dividing (1,1) and (7,7) in 1:2 are
   (a) (3,3)
   (b) (5,5)
   (c) (3,5)
   (d) (5,3)
   Answer: (a) – Explanation: Section formula ((17 +21)/3,(17 +21)/3)=(9/3,9/3)=(3,3); this midpoint-like calculation is key for finding centers in geometry problems.
4. The range of 8,3,11,1,6 is
   (a) 10
   (b) 8
   (c) 11
   (d) 5
   Answer: (a) – Explanation: Max 11 – min 1 =10; range gives spread of data, essential for understanding variability in statistics for students analyzing test scores.
5. The antiderivative of 2x dx is
   (a) x² + C
   (b) 2x² + C
   (c) x²/2 + C
   (d) 2x + C
   Answer: (a) – Explanation: ∫2x dx = x² + C; the power rule n x^{n-1} inverse is (x^n)/n, here n=2, divide by 2 but coefficient 2 cancels to x² + C, foundational for area under curves.
6. The value of cos 60° is
   (a) 1/2
   (b) √3/2
   (c) 0
   (d) –1/2
   Answer: (a) – Explanation: Cos 60° = adjacent/hypotenuse =1/2 in 30-60-90 triangle, often used in vector projections and cosine law for triangles.
7. The sum of roots for x² – 7x + 10 = 0 is
   (a) 7
   (b) –7
   (c) 10
   (d) –10
   Answer: (a) – Explanation: For ax² + bx + c =0, sum = –b/a =7/1=7; this Vieta’s formula saves time in quadratic problems without solving fully.
8. The section formula for points (2,3) and (8,9) in ratio 1:1 is
   (a) (5,6)
   (b) (6,5)
   (c) (4,6)
   (d) (6,4)
   Answer: (a) – Explanation: Midpoint ((2+8)/2,(3+9)/2)=(5,6); ratio 1:1 is midpoint, crucial for medians in triangles.
9. The mode of the data 4,5,5,6,6,6,7 is
   (a) 5
   (b) 6
   (c) 4
   (d) 7
   Answer: (b) – Explanation: 6 appears three times, highest frequency; mode helps identify most common value in data sets like survey results.
10. The identity sin²θ + cos²θ equals
    (a) 1
    (b) 0
    (c) tan²θ
    (d) sec²θ
    Answer: (a) – Explanation: Pythagorean theorem in unit circle, x=cos, y=sin, x² + y²=1; basis for many trig simplifications.
11. The slope of the line 3x + 4y = 12 is
    (a) –3/4
    (b) 3/4
    (c) 4/3
    (d) –4/3
    Answer: (a) – Explanation: y = –(3/4)x +3; slope m= –3/4, negative for downward line.
12. The probability of getting a head in a fair coin toss is
    (a) 1/2
    (b) 1
    (c) 0
    (d) 1/4
    Answer: (a) – Explanation: Two outcomes, one favorable; probability = favorable/total, fundamental for independent events.
13. The determinant of matrix [[1,2],[3,4]] is
    (a) –2
    (b) 2
    (c) –10
    (d) 10
    Answer: (a) – Explanation: ad – bc =4 –6 = –2; det used for inverse and solvability.
14. The 6th term of AP 2,5,8,… is
    (a) 17
    (b) 20
    (c) 23
    (d) 26
    Answer: (a) – Explanation: a=2, d=3, T6 =2 +5*3 =17; term formula a+(n–1)d for sequences.
15. The volume of a cube with side 4 cm is
    (a) 64 cm³
    (b) 16 cm³
    (c) 32 cm³
    (d) 8 cm³
    Answer: (a) – Explanation: V = a³ =64 cm³; basic for 3D shapes in mensuration.
16. The inverse of [[2,0],[0,3]] is
    (a) [[1/2,0],[0,1/3]]
    (b) [[0,2],[3,0]]
    (c) [[3,0],[0,2]]
    (d) [[1/2,3],[0,1/3]]
    Answer: (a) – Explanation: Diagonal, inverse reciprocals det=6, but simple 1/2,1/3; for diagonal matrices.
17. The number of ways to choose 2 students from 5 is
    (a) 10
    (b) 5
    (c) 25
    (d) 120
    Answer: (a) – Explanation: C(5,2)=10; combinations for selections without order.
18. The variance of 1,2,3,4 is
    (a) 1.25
    (b) 1
    (c) 1.5
    (d) 2
    Answer: (a) – Explanation: Mean=2.5; d²=2.25,0.25,0.25,2.25 sum5; variance=5/4=1.25.
19. The discriminant of x² + 2x + 1 =0 is
    (a) 0
    (b) 4
    (c) –4
    (d) 1
    Answer: (a) – Explanation: D=b²–4ac=4–4=0; equal roots, perfect square.
20. The length of tangent from (0,0) to x² + y² =25 is
    (a) 5
    (b) 0
    (c) 25
    (d) √25
    Answer: (b) – Explanation: Point on circle, tangent length 0; power =0.

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

Brief; steps shown with student tip.

Q21. Find the HCF of 54 and 72 using Euclid’s algorithm.
Answer: 72 =1×54 +18, 54 =3×18 +0. HCF=18.
Explanation: Euclid’s method repeatedly divides and takes remainder until 0; efficient for large numbers, tip: always divide larger by smaller first.

Q22. Find the coordinates of the point dividing the line segment joining (3,4) and (–1,2) in the ratio 2:1.
Answer: ((2(-1) +13)/3, (22 +14)/3) = (1/3,10/3).
Explanation: Section formula for internal division m:n = (m x2 + n x1)/(m+n); here m=2 n=1 from (3,4), tip: sign ratio direction.

Q23. Find the standard deviation for the data 2,4,6,8,10.
Answer: Mean =6; d² =16,4,0,4,16 sum40; variance =40/5=8, SD=√8=2√2≈2.82.
Explanation: SD measures spread; calculate absolute deviations squared, average, square root; tip: use calculator for large data, but show steps for board.

Q24. Find the derivative of y = x^3 + 2x with respect to x.
Answer: dy/dx =3x² +2.
Explanation: Power rule: d(x^n)/dx = n x^{n-1}; for x^3 =3x², 2x =2; tip: constant term 0, sum rule for polynomials.

Q25. Find the sum of first 5 terms of the GP 3,6,12,24,…
Answer: S5 =3(2^5 –1)/(2–1)=3(32–1)=93.
Explanation: GP sum S_n = a (r^n –1)/(r–1); r=2, a=3; tip: for r>1, formula works, verify T5=3*16=48, sum 3+6+12+24+48=93.

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

Short; diagrams where useful, with student tip.

Q26. Solve the quadratic equation x² – 8x + 15 = 0 by factorisation method.
Answer: x² – 8x + 15 = (x–3)(x–5)=0; x=3 or 5.
Explanation: Find two numbers sum –8, product 15: –3 and –5; multiply (x–3)(x–5)=x²–8x+15; roots by zero product; tip: for integer roots, factorisation fastest, verify by plugging x=3: 9–24+15=0.

Q27. Find the area of the triangle with vertices at (0,0), (6,0) and (0,8).
Answer: Area = (1/2) | (0(0–8) +6(8–0) +0(0–0)) | = (1/2)|0 +48 +0|=24 square units.
Explanation: Shoelace formula for coordinates: (1/2)|x1(y2–y3) +x2(y3–y1) +x3(y1–y2)|; or base 6, height 8, (1/2)baseheight=24; tip: shoelace for irregular, but right triangle simple, plot to visualize.

Q28. The mean of 7 numbers is 25. Find the total sum. If one number is 20, what is the mean of the remaining 6 numbers?
Answer: Total sum =7*25=175. Mean of remaining 6 = (175–20)/6 =155/6 ≈25.83.
Explanation: Mean = sum/n, so sum = mean * n; subtract one, divide by n–1; tip: useful for adjusting data sets in statistics, like test scores after removing outlier.

Q29. Find the derivative of y = e^x sin x with respect to x.
Answer: dy/dx = e^x sin x + e^x cos x = e^x (sin x + cos x).
Explanation: Product rule (u v)’ = u’ v + u v’; u=e^x u’=e^x, v=sin x v’=cos x; tip: exponential never zero, combines nicely, common in calculus applications like growth models.

Q30. Find the 9th term of the AP 4, 7, 10, …
Answer: a=4, d=3, T9 =4 +8*3 =4+24=28.
Explanation: General term T_n = a + (n–1)d; n=9, (9–1)=8 terms d; tip: for increasing sequences, easy arithmetic, verify T1=4, T2=7.

Q31. Draw the graph of the linear equation y = 2x – 3 for x from –2 to 2.
Answer: Points: x=–2 y= –7, x=–1 y= –5, x=0 y= –3, x=1 y= –1, x=2 y=1. Line with slope 2, y-intercept –3.
Explanation: Plot points on Cartesian plane, connect; slope rise/run=2/1; tip: two points define line, more for accuracy, useful for understanding linear functions in economics.

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

Long; proofs gargantuan, with student tips.

Q32. Prove that the diagonals of a parallelogram bisect each other.
Answer: In parallelogram ABCD, diagonals AC, BD intersect at O. Vectors: OA + OC = OB + OD (opposite sides equal/parallel). Midpoint of AC = (A+C)/2, midpoint BD = (B+D)/2. Since A + C = B + D, midpoints same O. Diagram: Parallelogram ABCD, diagonals cross at O, halves equal.
Explanation: Vector or coordinate proof (place A at (0,0), B (a,0), D (0,b), C (a,b); midpoint AC (a/2,b/2), BD (a/2,b/2)); tip: coordinate geometry simplifies, apply to rhombus/rectangle too.

Q33. Find the equation of the circle passing through the points (0,0), (4,0) and (0,3).
Answer: General x² + y² + Dx + Ey + F =0. Plug (0,0): F=0. (4,0): 16 +4D +F=0, D= –4. (0,3): 9 +3E +F=0, E= –3. x² + y² –4x –3y =0. Complete square: (x–2)² + (y–3/2)² =4 +9/4 =25/4, centre (2,1.5), r=5/2.
Explanation: Substitute three points to solve D,E,F; complete square for centre/radius; tip: three points determine unique circle, verify by plugging back.

Q34. (Choice: (a) or (b))
(a) Solve the system of equations 5x – 2y =4 and 3x + y =7 using substitution method.**
Answer: From second y=7–3x. Plug first 5x –2(7–3x)=4, 5x –14 +6x =4, 11x=18, x=18/11. y=7–3(18/11)= (77–54)/11=23/11. Explanation: Solve one for variable, substitute; verify 5(18/11) –2(23/11)= (90–46)/11=44/11=4, 3(18/11) +23/11= (54+23)/11=77/11=7; tip: fractions exact, decimal approx 1.636,2.091.

(b) [Elimination method for same.]

Q35. A solid metallic sphere of radius 7 cm is melted and recast into a hollow spherical shell of internal radius 5 cm. Find the thickness of the shell.
Answer: V_sphere =4/3 π 343 =1372π/3. Hollow V =4/3 π (R³ –5³) =1372π/3; R³ –125 =343, R³ =468, R=∛468≈7.76 cm. Thickness = R –5 ≈2.76 cm. Accurate: R = ∛(343 +125) =∛468. Simplify ∛(468)=∛(4117)=∛(49*13)=3∛(52/9)? But numerical.
Explanation: Volume conserved; outer R from V =4/3 π (R³ – r³); tip: cube root for exact, approximate for thickness, real-life casting problems.

Q36. If P(A) = 0.6, P(B) = 0.5 and P(A ∪ B) = 0.8, find P(A ∩ B).
Answer: P(A ∪ B) = P(A) + P(B) – P(A ∩ B); 0.8 =0.6 +0.5 – P; P=0.3.
Explanation: Union formula; intersection overlap; tip: if independent P(A)P(B)=0.3, but here dependent, useful for risk assessment in probability.

Q37. (Choice: (a) or (b))
(a) Prove that tan θ = sin θ / cos θ.**
Answer: In right triangle, tan = opposite/adjacent, sin = opposite/hyp, cos = adjacent/hyp; tan = sin/cos. Unit circle: y/x = (y/r) / (x/r) = sin/cos.
Explanation: Definition from triangle ratios; tip: memorize identities, derive from basics for exams.

(b) [Sec θ = 1/cos θ proof.]

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

Data-void; interpret entropy, with student tips.

Case 1 – Real Numbers (Unison World 2025)
Passage: Show √5 irrational. Assume p/q lowest.
(i) 5 q² = p², p divisible by 5, p=5k.
(ii) 5 q² =25 k², q²=5 k², q divisible by 5.
(iii) Contradiction, not lowest.
(iv) Hence irrational.
Explanation: Contradiction method; tip: apply to √n non-square, fundamental for number theory.

Case 2 – Linear Equations (Garhwal Public 2025)
Passage: 2x +3y =12, x + y =5. Solve.
(i) Subtract: x +2y =7.
(ii) From second x=5–y, plug 2(5–y) +3y =12, 10 –2y +3y =12, y=2, x=3.
(iii) Unique solution.
(iv) Graph intersecting.
Explanation: Elimination; tip: make coefficients same, subtract for variable.

Case 3 – Mensuration (Mussoorie International 2025)
Passage: Cylinder r=4 cm, h=10 cm. TSA? Volume?
(i) TSA =2πr h +2πr² =2π410 +2π16 =80π +32π=112π. (ii) V =π r² h =π1610=160π.
(iii) If melted to sphere, r_sphere from V=4/3 π r³ =160π, r³ =120, r=∛120≈4.93 cm.
(iv) Diagram cylinder.
Explanation: Formulas direct; tip: remember TSA includes bases, V only interior.

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

Q38. Blueprint: To verify the formula for the area of a trapezium using graph paper.
Aim: (a+b)/2 * h. Procedure: (1) Draw trapezium on grid with parallel sides a=6, b=4, h=5. (2) Count squares inside. (3) Area =5 squares = (6+4)/2 *5=25, but grid 1 unit square, adjust. Observation: Matches formula. Conclusion: Verified. Tip: Graph paper for irregular shapes. [Trapezium on grid diagram.]

Q39. Blueprint: To draw a pair of tangents to a circle from an external point.
Aim: Equal lengths. Procedure: (1) Circle centre O r=5. (2) Point P d=13. (3) Join OP, midpoint M, draw circle r= (d² – r²)^{1/2}/2? Standard: Construct perpendicular bisector or use compass to find touch points. Observation: Two tangents equal. Conclusion: Property holds. Tip: Symmetry. [Circle, P, two tangents diagram.]

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

1. Ur-Prep: 100% NCERT eclipse, 0% void; daily 2 hours practice.
2. Eclipse Exam Singularity: Pre: 10 min warm-up formulas; During: A singularity (10 min), B/C eon (25 min), D/E ur (60 min), 35 min eclipse review calculations.
3. Grimoire Art: Bold formulas; numbered steps; shaded diagrams for clarity.
4. Eclipse Scoring: 3-mark: 1 formula, 1 step, 1 answer; 5-mark: 5 clear points with tip.
5. Eclipse Shields: “Discriminant D=b²–4ac”; “AP T_n = a + (n–1)d”.
6. Case Eclipse: Read data twice, extract numbers, compute sequentially.
7. Practical Eclipse: Focus 5 constructions, 5 mensuration; practice with ruler.
8. Psyche Eclipse: Visualize perfect graph; if stuck, skip 1 min, return fresh.
9. Post-Eclipse: Score self, note 3 errors, next practice target fix.
10. Void-Eternal Sanctum: NCERT Exemplar for extras; YouTube for graph demos; group mocks weekly.