SEBA HSLC Mathematics Answer Key 2026 – Question Paper Solution & Detailed Explanation
The Assam HSLC Class 10th Maths Exam 2026 was conducted on 20 February 2026 under the supervision of the Board of Secondary Education Assam (SEBA). Students who appeared in the examination can now check the HSLC Maths Answer Key 2026 with detailed explanations to estimate their expected marks.
This article provides a subject-wise answer key, step-by-step solutions, and marking guidance to help students analyze their performance.
📅 Assam HSLC Maths Exam 2026 Overview
| Exam Name | Assam HSLC Class 10th Examination 2026 |
|---|---|
| Conducting Body | Board of Secondary Education Assam (SEBA) |
| Subject | Mathematics |
| Exam Date | 20 February 2026 |
| Answer Key Status | Unofficial (Subject Expert Prepared) |
| Official Result | To be announced |
SEBA HSLC Class 10th Maths Board Exam 2026
Complete Answer Key with Step-by-Step Explanations (Q1 to Q61)
PART A: Multiple Choice Questions (Q1 to Q45)
Q1. তলৰ কোনবোৰ দ্বিঘাত সমীকৰণৰ বাস্তৱ মূল নাই? (Which has no real roots?)
Answer: (c) (i) আৰু (iii)
Explanation: To find if roots are real, check the discriminant D = b² – 4ac. If D < 0, there are no real roots.
(i) D = (-2√3)² – 4(1)(9) = 12 – 36 = -24 < 0 (No real roots)
(ii) D = (-4√3)² – 4(3)(4) = 48 – 48 = 0 (Real roots exist)
(iii) D = (1)² – 4(1)(1) = 1 – 4 = -3 < 0 (No real roots)
Thus, (i) and (iii) have no real roots.
(i) D = (-2√3)² – 4(1)(9) = 12 – 36 = -24 < 0 (No real roots)
(ii) D = (-4√3)² – 4(3)(4) = 48 – 48 = 0 (Real roots exist)
(iii) D = (1)² – 4(1)(1) = 1 – 4 = -3 < 0 (No real roots)
Thus, (i) and (iii) have no real roots.
Q2. তলৰ কোনটো সমান্তৰ প্ৰগতি? (Which sequence is an AP?)
Answer: (b) -6, -4, -2, 0, …
Explanation: An Arithmetic Progression (AP) must have a constant common difference (d). In option (b), -4 – (-6) = 2 and -2 – (-4) = 2. The difference is consistently 2.
Q3. চিত্ৰত ΔABC আৰু ΔXYZ সদৃশ হ’লে, ∠X ৰ মান হ’ব
Answer: (b) 50°
Explanation: In ΔABC, ∠C = 180° – (60°+70°) = 50°. The sides of ΔXYZ are exactly double the sides of ΔABC (4.2 → 8.4, 7 → 14, 3√3 → 6√3). By SSS similarity, ΔABC ∼ ΔYZX. Angle X corresponds to Angle C. Therefore, ∠X = 50°.
Q4. যদি m আৰু n দুটা ধনাত্মক অখণ্ড সংখ্যা m = pq³ আৰু n = p³q², তেন্তে ল.সা.গু (LCM) × গ.সা.উ (HCF) = ?
Answer: (b) p⁴q⁵
Explanation: The formula states: LCM × HCF = Product of the numbers. Therefore, m × n = (pq³) × (p³q²) = p⁴q⁵.
Q5. এটা ৰৈখিক বহুপদ px – q ৰ শূন্য হ’ব
Answer: (c) q/p
Explanation: To find the zero of the polynomial, equate it to zero: px – q = 0 ⇒ px = q ⇒ x = q/p.
Q6. চিত্ৰত যদি ΔABC ৰ পৰিসীমা 27 cm, তেন্তে AP + BQ + CR ৰ মান হ’ব
Answer: (c) 13.5 cm
Explanation: Tangents drawn from an external point to a circle are equal (AP=AR, BP=BQ, CQ=CR). The perimeter of ΔABC is AB+BC+CA = 2(AP+BQ+CR) = 27 cm. Therefore, AP+BQ+CR = 27 / 2 = 13.5 cm.
Q7. 5 cm ব্যাসাৰ্ধৰ বৃত্ত এটাৰ পৰিধি হ’ব
Answer: (c) 31.4 cm
Explanation: Circumference = 2πr = 2 × 3.14 × 5 = 31.4 cm.
Q8. যদি P(A) কোনো ঘটনা A ৰ সম্ভাৱিতা হয় তেন্তে P(A) হ’ব
Answer: (d) 0 ≤ P(A) ≤ 1
Explanation: The probability of any event cannot be negative and cannot exceed 100% (or 1).
Q9. A(0, 0), B(6, 0) আৰু C(0, 8) এটা ত্ৰিভুজৰ শীৰ্ষবিন্দু। বাহুকেইডালৰ জোখৰ উৰ্ধ্বক্ৰম হ’ব
Answer: (d) AB < AC < BC
Explanation: Using the distance formula:
AB = √(6-0)² = 6
BC = √((0-6)² + (8-0)²) = √(36+64) = 10
AC = √((0-0)² + (8-0)²) = 8
In ascending order: 6 < 8 < 10 ⇒ AB < AC < BC.
AB = √(6-0)² = 6
BC = √((0-6)² + (8-0)²) = √(36+64) = 10
AC = √((0-0)² + (8-0)²) = 8
In ascending order: 6 < 8 < 10 ⇒ AB < AC < BC.
Q10. যদি cot θ = 7/8, তেন্তে (1 + sin θ)(1 – sin θ) / (1 + cos θ)(1 – cos θ) ৰ মান হ’ব
Answer: (d) 49/64
Explanation: The expression simplifies using (a+b)(a-b) = a²-b²:
(1 – sin²θ) / (1 – cos²θ) = cos²θ / sin²θ = cot²θ.
Since cotθ = 7/8, squaring it gives 49/64.
(1 – sin²θ) / (1 – cos²θ) = cos²θ / sin²θ = cot²θ.
Since cotθ = 7/8, squaring it gives 49/64.
Q11. কেন্দ্ৰৰ পৰা 3 cm দূৰত্বত থকা এডাল জ্যাৰ দৈৰ্ঘ্য 8 cm হ’লে বৃত্তৰ ব্যাসাৰ্ধৰ মান হ’ব
Answer: (d) 5 cm
Explanation: A perpendicular from the center bisects the chord. This creates a right-angled triangle with base 4 cm (half of 8) and height 3 cm. By Pythagoras theorem: r = √(3² + 4²) = √25 = 5 cm.
Q12. উক্তি (A): যদি বহুলক 60 আৰু মাধ্য 66 হয় তেন্তে মধ্যমা হ’ব 64.
যুক্তি (R): মধ্যমা = 1/3 (বহুলক + 2 × মাধ্য)
যুক্তি (R): মধ্যমা = 1/3 (বহুলক + 2 × মাধ্য)
Answer: (a) উক্তি (A) আৰু যুক্তি (R) দুয়োটাই শুদ্ধ আৰু (R) (A) ৰ শুদ্ধ ব্যাখ্যা।
Explanation: Using the empirical formula: 3 × Median = Mode + 2 × Mean ⇒ 3 × Median = 60 + 2(66) = 192 ⇒ Median = 64. Both statements are correct and the reason explains the assertion perfectly.
Q13. শংকুৰ মাত্ৰাৰ স্তম্ভ মিলাওক (Match the following for Cone dimensions):
Answer: (c) (i) → (Q), (ii) → (P)
Explanation: For a cone with diameter d, radius r = d/2.
Curved Surface Area (CSA) = πrl = π(d/2)l = πdl/2 (Q).
Base Area = πr² = π(d/2)² = πd²/4 (P).
Curved Surface Area (CSA) = πrl = π(d/2)l = πdl/2 (Q).
Base Area = πr² = π(d/2)² = πd²/4 (P).
Q14. যদি kx – 5y = 2, 6x + 2y = 7 সমীকৰণযোৰৰ কোনো সমাধান নাথাকে তেন্তে k ৰ মান হ’ব
Answer: (d) -15
Explanation: For no solution, the lines must be parallel: a₁/a₂ = b₁/b₂ ≠ c₁/c₂ ⇒ k/6 = -5/2 ⇒ 2k = -30 ⇒ k = -15.
Q15. যদি 4, x₁, x₂, x₃, 28 সমান্তৰ প্ৰগতিত থাকে, তেন্তে x₃ ৰ মান হ’ব
Answer: (c) 22
Explanation: First term a = 4. The 5th term is 28 ⇒ a + 4d = 28 ⇒ 4 + 4d = 28 ⇒ d = 6.
The term x₃ is the 4th term, which is a + 3d = 4 + 3(6) = 22.
The term x₃ is the 4th term, which is a + 3d = 4 + 3(6) = 22.
Q16. যদি α আৰু β, বহুপদ px² + qx + r ৰ দুটা শূন্য হয়, তেন্তে (1 + α)(1 + β) ৰ মান হ’ব
Answer: (d) (p – q + r) / p
Explanation: Expanding (1+α)(1+β) = 1 + (α+β) + αβ.
Sum of zeroes (α+β) = -q/p, Product (αβ) = r/p.
Substituting: 1 – q/p + r/p = (p – q + r) / p.
Sum of zeroes (α+β) = -q/p, Product (αβ) = r/p.
Substituting: 1 – q/p + r/p = (p – q + r) / p.
Q17. তলৰ কোনটো ৰৈখিক সমীকৰণ যোৰ অসঙ্গত (Inconsistent)?
Answer: (c) কেৱল (i) অসঙ্গত
Explanation: Inconsistent equations represent parallel lines (a₁/a₂ = b₁/b₂ ≠ c₁/c₂). In (i): 2x – 3y = 8 and 4x – 6y = 9, 2/4 = -3/-6 ≠ 8/9, which fulfills this condition.
Q18. সমীকৰণৰ মূলৰ প্ৰকৃতি মিলাওক (Match the nature of roots):
Answer: (b) P → (ii), Q → (i), R → (iv), S → (iii)
Explanation: Solving the discriminant:
(Q) x²-2x+1=0 ⇒ D = 4-4=0 (Real and equal roots).
(R) x²-x+6=0 ⇒ D = 1-24 = -23 < 0 (Imaginary roots).
(Q) x²-2x+1=0 ⇒ D = 4-4=0 (Real and equal roots).
(R) x²-x+6=0 ⇒ D = 1-24 = -23 < 0 (Imaginary roots).
Q19. 3087 ক আটাইতকৈ সৰু কি সংখ্যাৰে পূৰণ কৰিলে সংখ্যাটো এটা পূৰ্ণবৰ্গ হ’ব?
Answer: (b) 7
Explanation: Prime factorization of 3087 = 3² × 7³. To make it a perfect square (where all exponents are even), we must multiply by 7 to get 3² × 7⁴.
Q20. তলৰ কোনটো এটা অপৰিমেয় সংখ্যা নহয়? (Which is NOT an irrational number?)
Answer: (c) (√2-√3)(√2+√3)
Explanation: Expanding gives (√2)² – (√3)² = 2 – 3 = -1. Because -1 is an integer (rational number), it is NOT irrational.
Q21. (2, -1) আৰু (6, 3) বিন্দু সংযোগী ৰেখাখণ্ডৰ মধ্যবিন্দুৰ স্থানাংক হ’ব
Answer: (a) (4, 1)
Explanation: Midpoint formula M = ((x₁+x₂)/2, (y₁+y₂)/2) = ((2+6)/2, (-1+3)/2) = (8/2, 2/2) = (4, 1).
Q22. যদি sin θ + cos θ = 4/3, তেন্তে sin θ · cos θ ৰ মান হ’ব
Answer: (c) 7/18
Explanation: Square both sides of sinθ + cosθ = 4/3.
sin²θ + cos²θ + 2sinθcosθ = 16/9
1 + 2sinθcosθ = 16/9 ⇒ 2sinθcosθ = 7/9 ⇒ sinθcosθ = 7/18.
sin²θ + cos²θ + 2sinθcosθ = 16/9
1 + 2sinθcosθ = 16/9 ⇒ 2sinθcosθ = 7/9 ⇒ sinθcosθ = 7/18.
Q23. O কেন্দ্ৰীয় বৃত্তত PA আৰু PB স্পৰ্শক আৰু ∠APB = 50°, তেন্তে ∠OAB ৰ মান হ’ব
Answer: (a) (i), (iv)
Explanation: In quadrilateral OAPB, ∠AOB = 180° – 50° = 130°. In isosceles ΔOAB (since OA=OB), ∠OAB = (180° – 130°)/2 = 25°. Note that option (iv) 90°-65° = 25° is also mathematically identical.
Q24. এটা সমান্তৰ প্ৰগতিৰ প্ৰথম পদ ‘a’ আৰু সাধাৰণ অন্তৰ ‘p’, তেন্তে ইয়াৰ n তম পদটো হ’ব
Answer: (d) a + (n-1)p
Explanation: This is the standard n-th term formula for an AP with common difference p instead of d.
Q25. দুটা সদৃশ ত্ৰিভুজৰ ক্ষেত্ৰত সদায়েই সত্য হ’ব?
Answer: (c) সিহঁতৰ অনুৰূপ কোণবোৰ সমান (Corresponding angles are equal)
Explanation: Similar triangles always have equal corresponding angles and proportional (not necessarily equal) corresponding sides.
Q26. এটা গোলকৰ আয়তন আৰু পৃষ্ঠকালি সাংখ্যিকভাৱে সমান। গোলকটোৰ ব্যাস হ’ব
Answer: (b) 6 একক (units)
Explanation: Volume = Surface Area ⇒ (4/3)πr³ = 4πr² ⇒ r = 3. The diameter d = 2r = 6.
Q27. প্ৰথম 8 টা মৌলিক সংখ্যাৰ মধ্যমা হ’ল
Answer: (b) 9
Explanation: The first 8 primes are 2, 3, 5, 7, 11, 13, 17, 19. The median is the average of the 4th and 5th terms: (7+11)/2 = 9.
Q28. কোনো ঘটনা E ৰ সম্ভাৱিতা P(E) = 0.11 হ’লে P(Not E) ৰ মান হ’ব
Answer: (c) 0.89
Explanation: P(Not E) = 1 – P(E) = 1 – 0.11 = 0.89.
Q29. এডাল ৰেখা AB ক 3:4 অনুপাতত ভাগ কৰোঁতে অংকনৰ পৰ্যায়বোৰৰ ক্ৰম হ’ল:
Answer: (a) (vi), (i), (iv), (iii), (ii), (v)
Explanation: Chronological steps: Given a line segment (vi), draw an acute angle at A (i), plot 7 points (iv), join the last point to B (iii), draw a parallel line (ii), which divides the line (v).
Q30. দুটা বৃত্তৰ ব্যাসাৰ্ধ 6 cm আৰু 8 cm। বৃত্ত দুটাৰ পৰিধিৰ সমষ্টিৰ সমান হোৱা নতুন বৃত্তটোৰ ব্যাস হ’ব
Answer: (d) 14 cm (Note: Board likely asked for radius based on options)
Explanation: 2πR = 2π(6) + 2π(8) ⇒ R = 6 + 8 = 14 cm. The value 14 corresponds to the radius.
Q31. উক্তি (A): বহুপদ 3x² + kx + 5 ৰ শূন্য দুটাৰ যোগফল 2/3, তেন্তে k = -2। যুক্তি (R): ax² + bx + c ৰ শূন্য দুটাৰ পূৰণফল c/a।
Answer: (b) দুয়োটাই সত্য, কিন্তু (R), (A) ৰ শুদ্ধ ব্যাখ্যা নহয়।
Explanation: Sum of zeroes is -b/a. So, -k/3 = 2/3 ⇒ k = -2 (A is True). Reason R states the product rule, which is true, but it doesn’t explain the sum rule used in Assertion A.
Q32. উক্তি (A): সকলো বৰ্গই সদৃশ। উক্তি (B): সমান সংখ্যক বাহুৰ দুটা বহুভুজ সদৃশ হ’ব যদি সিহঁতৰ অনুৰূপ বাহুবোৰ সমান হয়।
Answer: (a) (A) সত্য কিন্তু (B) অসত্য।
Explanation: (A) All squares are similar (True). (B) Polygons are similar if corresponding angles are equal and sides are *proportional*, not necessarily equal (False).
Q33. ax² + bx + c = 0 সমীকৰণটো সিদ্ধ কৰা x ৰ মান হ’ব
Answer: (c) -b/2a ± √(b²-4ac) / 2a
Explanation: This is the standard quadratic formula written with split denominators.
Q34. n² আৰু (n + 1)² ৰ মাজত কিমানটো স্বাভাৱিক সংখ্যা আছে?
Answer: (a) 2n
Explanation: There are exactly 2n non-perfect square natural numbers strictly between n² and (n+1)².
Q35. উক্তি (I): 7 × 2 + 3 এটা যৌগিক সংখ্যা। উক্তি (II): প্ৰত্যেক যৌগিক সংখ্যাকে মৌলিক সংখ্যাৰ ঘাতৰ গুণফল ৰূপে লিখিব পাৰি।
Answer: (d) (I) অসত্য কিন্তু (II) সত্য
Explanation: 7 × 2 + 3 = 17, which is a prime number, not composite. So (I) is false. (II) is the Fundamental Theorem of Arithmetic, which is true.
Q36. (-4, 6) আৰু (8, -6) বিন্দু দুটাৰ পৰা সমদূৰৱৰ্তী x-অক্ষৰ ওপৰত থকা বিন্দুটোৰ স্থানাংক হ’ব
Answer: (c) (2, 0)
Explanation: Let point on x-axis be (x,0). Using the distance formula:
(x+4)² + 36 = (x-8)² + 36 ⇒ x²+8x+16 = x²-16x+64 ⇒ 24x = 48 ⇒ x = 2.
(x+4)² + 36 = (x-8)² + 36 ⇒ x²+8x+16 = x²-16x+64 ⇒ 24x = 48 ⇒ x = 2.
Q37. উক্তি (A): tan A ৰ মান সদায় 1 তকৈ সৰু। উক্তি (B): A কোণৰ কোনো মানৰ বাবে sec A = 12/5
Answer: (b) (A) অসত্য, (B) সত্য।
Explanation: (A) tan A can be greater than 1 (e.g., tan 60° ≈ 1.732). (B) sec A ≥ 1 always, so 12/5 = 2.4 is a valid value.
Q38. উক্তি (A): বৃত্ত এটাত অসীম সংখ্যক সমান্তৰাল স্পৰ্শক টানিব পাৰি। উক্তি (B): যদি ABCD চতুৰ্ভুজৰ বাহুকেইটাই এটা বৃত্তক স্পৰ্শ কৰে তেন্তে AB + CD = BC + AD
Answer: (c) (A) অসত্য কিন্তু (B) সত্য।
Explanation: (A) A circle can have at most *two* parallel tangents corresponding to a secant. (B) This is a standard theorem for a circumscribing quadrilateral.
Q39. যদি দুডাল ৰেখা সমান্তৰাল হয় আৰু এডাল ৰেখাৰ সমীকৰণ 2x – 3y = 5 হয়, তেন্তে আনডাল ৰেখাৰ সমীকৰণ হ’ব পাৰে
Answer: (b) 4x – 6y = 15
Explanation: For parallel lines, coefficients of x and y must be proportional. 2/4 = -3/-6 = 1/2. This ratio is not equal to 5/15, perfectly satisfying parallel line conditions.
Q40. √2, √8, √18, √32 সমান্তৰ প্ৰগতিটোৰ পৰৱৰ্তী পদটো হ’ব
Answer: (b) √50
Explanation: The sequence simplifies to √2, 2√2, 3√2, 4√2. The next term is 5√2 = √(25 × 2) = √50.
Q41. প্ৰদত্ত বিভাজন তালিকাৰ বহুলক শ্ৰেণী হ’ব (Modal Class)
Answer: (c) 55-70
Explanation: The modal class is the class with the highest frequency. The highest frequency here is 8, which corresponds to the class interval 55-70.
Q42. উক্তি (A): দুটা মুদ্ৰা একেলগে টছ কৰিলে অতি কমেও এটা মুণ্ড পোৱাৰ সম্ভাৱিতা 1/2। উক্তি (B): 52 টা কাৰ্ডৰ পৰা এটা ৰজা হোৱাৰ সম্ভাৱিতা 1/13।
Answer: (b) (A) অশুদ্ধ, (B) শুদ্ধ।
Explanation: (A) Probability of at least one head in two tosses (HH, HT, TH) is 3/4, not 1/2. (B) Probability of drawing a king is 4/52 = 1/13.
Q43. x² – y² + x + y ৰ এটা উৎপাদক (x + y) হ’লে আনটো উৎপাদক হ’ব
Answer: (a) (1 + x – y)
Explanation: Grouping and factoring: x² – y² + x + y = (x-y)(x+y) + 1(x+y) = (x+y)(x – y + 1). The other factor is (1 + x – y).
Q44. মুখ্য বৃত্তকলাৰ কালি হ’ব (Area of major sector)
Answer: (c) (i), (iv)
Explanation: The area of a major sector is ((360° – θ)/360°) × πr². Option (i) mathematically simplifies to this exact formula ((360-θ)/720) × 2πr² by cancelling the 2.
Q45. দুটা গোলকৰ আয়তনৰ অনুপাত 125:8 হ’লে, সিহঁতৰ পৃষ্ঠকালিৰ অনুপাত হ’ব
Answer: (b) 25:4
Explanation: Ratio of volumes is (r₁/r₂)³ = 125/8 ⇒ r₁/r₂ = 5/2. Ratio of surface areas is (r₁/r₂)² = (5/2)² = 25/4.
PART B: Descriptive Questions (Q46 to Q61)
Q46. 1620 ক আটাইতকৈ সৰু কি সংখ্যাৰে হৰণ কৰিলে সংখ্যাটো এটা পূৰ্ণবৰ্গ হ’ব?
Answer: 5
Explanation: Prime factorizing 1620 gives 1620 = 2² × 3⁴ × 5. To make all exponents even (a requirement for perfect squares), we must divide the number by the unpaired factor, which is 5.
Q47. মৌলিক উৎপাদকীকৰণ পদ্ধতিৰে 96 আৰু 404 ৰ গ.সা.উ. (HCF) আৰু ল.সা.গু. (LCM) উলিওৱা।
Answer: HCF = 4, LCM = 9696
Explanation:
96 = 2⁵ × 3
404 = 2² × 101
HCF (lowest power of common factors) = 2² = 4.
LCM = (96 × 404) / HCF = 38784 / 4 = 9696.
96 = 2⁵ × 3
404 = 2² × 101
HCF (lowest power of common factors) = 2² = 4.
LCM = (96 × 404) / HCF = 38784 / 4 = 9696.
Q48. প্ৰমাণ কৰা যে: (cosec θ – cot θ)² = (sec θ – 1) / (sec θ + 1)
Answer: Proved LHS = RHS
Explanation (Proof Steps):
LHS = (1/sinθ – cosθ/sinθ)²
= (1 – cosθ)² / sin²θ
= (1 – cosθ)² / (1 – cos²θ)
= ((1 – cosθ)(1 – cosθ)) / ((1 – cosθ)(1 + cosθ))
= (1 – cosθ) / (1 + cosθ)
Divide numerator and denominator by cosθ:
= (1/cosθ – 1) / (1/cosθ + 1) = (secθ – 1) / (secθ + 1) = RHS.
LHS = (1/sinθ – cosθ/sinθ)²
= (1 – cosθ)² / sin²θ
= (1 – cosθ)² / (1 – cos²θ)
= ((1 – cosθ)(1 – cosθ)) / ((1 – cosθ)(1 + cosθ))
= (1 – cosθ) / (1 + cosθ)
Divide numerator and denominator by cosθ:
= (1/cosθ – 1) / (1/cosθ + 1) = (secθ – 1) / (secθ + 1) = RHS.
Q49. মান নিৰ্ণয় কৰা: (2tan 30°) / (1 + tan² 30°)
Answer: √3 / 2
Explanation: Substitute tan 30° = 1/√3.
= (2/√3) / (1 + 1/3)
= (2/√3) / (4/3)
= (2/√3) × (3/4) = √3 / 2.
= (2/√3) / (1 + 1/3)
= (2/√3) / (4/3)
= (2/√3) × (3/4) = √3 / 2.
Q50. ভালদৰে মিহলোৱা 52 টা কাৰ্ডৰ পৰা এটা কাৰ্ড টানিলে সম্ভাৱিতা উলিওৱা: (i) এটা মুখ কাৰ্ড, (ii) এটা ইস্কাপন।
Answer: (i) 3/13, (ii) 1/4
Explanation:
(i) Face Card: There are 12 face cards in a deck. P = 12/52 = 3/13.
(ii) Spade: There are 13 Spades in a deck. P = 13/52 = 1/4.
(i) Face Card: There are 12 face cards in a deck. P = 12/52 = 3/13.
(ii) Spade: There are 13 Spades in a deck. P = 13/52 = 1/4.
Q51. 3x³ – x² – 3x + 1 বহুপদৰ এটা শূন্য -1 হ’লে বাকী কেইটা শূন্য নিৰ্ণয় কৰা।
Answer: 1 আৰু 1/3
Explanation: Since -1 is a zero, (x+1) is a factor. Divide the polynomial by (x+1):
Quotient = 3x² – 4x + 1.
Now factorize: 3x² – 3x – x + 1 = 0 ⇒ (3x-1)(x-1) = 0.
The other zeroes are x = 1 and x = 1/3.
Quotient = 3x² – 4x + 1.
Now factorize: 3x² – 3x – x + 1 = 0 ⇒ (3x-1)(x-1) = 0.
The other zeroes are x = 1 and x = 1/3.
Q52. দুটা অংক বিশিষ্ট এটা সংখ্যা আৰু সেই সংখ্যাটোৰ অংক দুটা সলসলনি কৰি পোৱা সংখ্যাটো যোগ কৰিলে 66 হয়। অংক দুটাৰ পাৰ্থক্য 2 হ’লে সংখ্যাটো নিৰ্ণয় কৰা।
Answer: 42 বা 24
Explanation: Let the number be 10x+y. Reversed is 10y+x.
Sum: (10x+y) + (10y+x) = 66 ⇒ 11(x+y) = 66 ⇒ x+y = 6.
Difference: x-y = 2 OR y-x = 2.
Solving x+y=6 and x-y=2 gives x=4, y=2 (Number is 42).
Solving x+y=6 and y-x=2 gives x=2, y=4 (Number is 24).
Sum: (10x+y) + (10y+x) = 66 ⇒ 11(x+y) = 66 ⇒ x+y = 6.
Difference: x-y = 2 OR y-x = 2.
Solving x+y=6 and x-y=2 gives x=4, y=2 (Number is 42).
Solving x+y=6 and y-x=2 gives x=2, y=4 (Number is 24).
Q53. মূল নিৰ্ণয় কৰা: 132/x – 1 = 132/(x+11)
Answer: 33 আৰু -44
Explanation:
132/x – 132/(x+11) = 1
132 = 1
132 × 11 = x² + 11x ⇒ x² + 11x – 1452 = 0.
Factorizing: (x+44)(x-33) = 0. Roots are x = 33 and x = -44.
132/x – 132/(x+11) = 1
132 = 1
132 × 11 = x² + 11x ⇒ x² + 11x – 1452 = 0.
Factorizing: (x+44)(x-33) = 0. Roots are x = 33 and x = -44.
Q54. ত্ৰেপিজিয়াম PQRS ত PQ || RS. M আৰু N ক্ৰমে PS আৰু QR ৰ ওপৰত… প্ৰমাণ কৰা যে MN || PQ.
Answer: Proved using Basic Proportionality Theorem (Thales Theorem).
Explanation: Draw a diagonal PR intersecting MN at point O. In ΔPSR, since PM/MS corresponds to the split on the diagonal, MO || RS. Similarly, in ΔPQR, ON || PQ. Since RS || PQ, therefore MN is parallel to PQ.
Q55. A(-2, -2) আৰু B(2, -4) সংযোগী ৰেখাখণ্ডত P বিন্দুৰ স্থানাংক নিৰ্ণয় কৰা যাতে PB = (2/5)AB.
Answer: (2/5, -16/5)
Explanation: If PB = (2/5)AB, then AP = (3/5)AB. P divides line AB internally in the ratio 3:2.
Section Formula: x = / 5 = 2/5.
y = / 5 = -16/5.
Section Formula: x = / 5 = 2/5.
y = / 5 = -16/5.
Q56. বহিঃবিন্দু T ৰ পৰা O কেন্দ্ৰীয় বৃত্তলৈ দুডাল স্পৰ্শক TP আৰু TQ টনা হ’ল। প্ৰমাণ কৰা যে ∠PTQ = 2∠OPQ.
Answer: Proved geometrically.
Explanation: Let ∠PTQ = θ. Since TP = TQ, ΔPTQ is isosceles.
∠TPQ = (180° – θ)/2 = 90° – θ/2.
Radius OP is perpendicular to tangent TP, so ∠OPT = 90°.
∠OPQ = ∠OPT – ∠TPQ = 90° – (90° – θ/2) = θ/2.
Hence, θ = 2∠OPQ ⇒ ∠PTQ = 2∠OPQ.
∠TPQ = (180° – θ)/2 = 90° – θ/2.
Radius OP is perpendicular to tangent TP, so ∠OPT = 90°.
∠OPQ = ∠OPT – ∠TPQ = 90° – (90° – θ/2) = θ/2.
Hence, θ = 2∠OPQ ⇒ ∠PTQ = 2∠OPQ.
Q57. 21 cm ব্যাসাৰ্ধৰ বৃত্তত কেন্দ্ৰত 60° কোণ কৰিলে নিৰ্ণয় কৰা: চৰ্পৰ দৈৰ্ঘ্য, পৰিসীমা আৰু কালি।
Answer: (i) 22 cm, (ii) 64 cm, (iii) 231 cm²
Explanation:
(i) Arc Length = (60/360) × 2 × (22/7) × 21 = 22 cm.
(ii) Perimeter = Arc Length + 2r = 22 + 42 = 64 cm.
(iii) Area of Sector = (60/360) × (22/7) × 21² = 231 cm².
(i) Arc Length = (60/360) × 2 × (22/7) × 21 = 22 cm.
(ii) Perimeter = Arc Length + 2r = 22 + 42 = 64 cm.
(iii) Area of Sector = (60/360) × (22/7) × 21² = 231 cm².
Q58. এটা লোহাৰ খুঁটাৰ আয়তন নিৰ্ণয় কৰা (Cylinder + Cone combination).
Answer: 34,496 cm³
Explanation: Radius r = 7 cm. Cone height h₁ = 42 cm. Cylinder height h₂ = 252 – 42 = 210 cm.
Total Vol = πr²h₂ + (1/3)πr²h₁ = πr²(h₂ + h₁/3)
= (22/7) × 49 × (210 + 14) = 154 × 224 = 34,496 cm³.
Total Vol = πr²h₂ + (1/3)πr²h₁ = πr²(h₂ + h₁/3)
= (22/7) × 49 × (210 + 14) = 154 × 224 = 34,496 cm³.
Q59. নিৰ্ণয় কৰা x আৰু y ৰ মান, যদি N = 60 আৰু মধ্যমা 28.5।
Answer: x = 5, y = 15
Explanation: Sum of frequencies = x + y + 40 = 60 ⇒ x + y = 20.
Median 28.5 lies in class 20-30. L = 20, f = 20, CF = x + 8, h = 10.
Formula: 28.5 = 20 + × 10
8.5 = (22 – x) / 2 ⇒ 17 = 22 – x ⇒ x = 5.
Since x + y = 20, then y = 15.
Median 28.5 lies in class 20-30. L = 20, f = 20, CF = x + 8, h = 10.
Formula: 28.5 = 20 + × 10
8.5 = (22 – x) / 2 ⇒ 17 = 22 – x ⇒ x = 5.
Since x + y = 20, then y = 15.
Q60. 7 cm, 6 cm আৰু 4 cm বাহুযুক্ত এটা ত্ৰিভুজ অঁকা আৰু তাৰ পিছত ইয়াৰ লগত সদৃশ হোৱাকৈ আন এটা ত্ৰিভুজ অঁকা যাৰ বাহুবোৰ প্ৰথম ত্ৰিভুজটোৰ 5/3 গুণ।
Answer: Geometrical Construction.
Explanation (Steps):
1. Draw base 7 cm. Construct the original triangle using 6cm and 4cm arcs.
2. Draw an acute angle ray below the base and mark 5 equal segments (since 5 is the larger number in 5/3).
3. Join the 3rd mark (denominator) to the end of the base.
4. Draw a parallel line from the 5th mark to intersect the extended base.
5. Draw a parallel line to the original side from this new point to complete the enlarged triangle.
2. Draw an acute angle ray below the base and mark 5 equal segments (since 5 is the larger number in 5/3).
3. Join the 3rd mark (denominator) to the end of the base.
4. Draw a parallel line from the 5th mark to intersect the extended base.
5. Draw a parallel line to the original side from this new point to complete the enlarged triangle.
Q61. গছপুলি ৰোপণ কাৰ্যসূচী… (Tree Plantation AP Case Study)
Answer: (i) 110, (ii) 30, (iii) 12 rows
Explanation: This forms an Arithmetic Progression (AP) where a = 20, d = 10.
(i) 10th row: a₁₀ = a + 9d = 20 + 90 = 110.
(ii) Difference (6th and 3rd): 3d = 30.
(iii) Total 900 trees: Sₙ = (n/2) = 900
5n² + 15n – 900 = 0 ⇒ n² + 3n – 180 = 0
(n+15)(n-12) = 0. Since n must be positive, n = 12 rows.
(i) 10th row: a₁₀ = a + 9d = 20 + 90 = 110.
(ii) Difference (6th and 3rd): 3d = 30.
(iii) Total 900 trees: Sₙ = (n/2) = 900
5n² + 15n – 900 = 0 ⇒ n² + 3n – 180 = 0
(n+15)(n-12) = 0. Since n must be positive, n = 12 rows.
📝 Assam HSLC Maths Answer Key 2026 (20 February)
Below is the unofficial answer key with explanation prepared by subject experts based on memory-based questions.
📊 Expected Cut-Off & Marks Analysis
- Easy to Moderate Difficulty
- MCQs were scoring
- Algebra & Trigonometry had direct formula-based questions
- Expected good score range: 60–85+ marks
📥 How to Calculate Your Expected Marks?
- Match your answers with the answer key.
- Follow SEBA marking scheme.
- Add marks for correct answers.
- Avoid negative marking (if applicable).
Disclaimer
This answer key and solution set is provided for educational and reference purposes only. While every effort has been made to ensure accuracy and provide detailed step-by-step solutions based on the SEBA HSLC Class 10th Maths Exam 2026 question paper, this is an unofficial answer key prepared by subject experts.
The official Board of Secondary Education, Assam (SEBA) marking scheme and final evaluation guidelines may vary slightly. Students are strongly advised to cross-check with their school teachers or official SEBA notifications for final confirmation. The website/author assumes no liability for any errors, omissions, or any decisions made based on this information.
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