10th Maths Half Yearly 2025 Answer Key | PDF Download Link
Are you searching for the original 10th Maths Half Yearly Question Paper 2025 with a verified answer key? You've come to the right place. This guide provides a detailed exam pattern analysis, chapter-wise weightage, expert strategies, and essential resources to help you ace your Tamil Nadu half-yearly exams.
📊 Overview of the 10th Maths Half Yearly Exam 2025 Pattern (Expected)
The Tamil Nadu State Board 10th Maths Half Yearly Exam typically covers the first 4-5 chapters and is designed to test conceptual understanding, problem-solving speed, and application skills.
Expected Key Sections & Mark Distribution:
Multiple Choice Questions (1 Mark each): 1-14 questions
Short Answer Questions (2 Marks each): 15-28 questions
Long Answer Questions I (5 Marks each): 29-42 questions
Long Answer Questions II (8 Marks each): 43-44 questions
Expected Chapter-Wise Weightage:
Relations and Functions & Numbers and Sequences: ~25 Marks
Algebra: ~35 Marks
Geometry: ~20 Marks
Statistics & Probability: ~10 Marks
🧠 Expert Tips to Score 90+ in Maths Half Yearly Exam
Master the Theorems and Formulas (30% of your success): Create a separate formula sheet for all chapters. Understanding the derivation of formulas helps in remembering and applying them correctly.
Focus on Problem-Solving, Not Rote Learning: Maths is about application. For each concept (e.g., "Euclid's Division Lemma," "Graph of a Quadratic Equation"), practice at least 5-7 problems of varying difficulty.
Time Management is Critical:
First 15 mins: Skim the paper. Solve all 1-mark questions.
Next 50 mins: Tackle the 2-mark and 5-mark questions you are most confident about.
Last 40 mins: Devote time to the challenging 8-mark problems (like geometry proofs or word problems from Algebra).
Final 15 mins: Review your paper, especially calculations and steps in long answers.
Step-by-Step Solutions are Mandatory: Even if the final answer is wrong, you get marks for the correct method. Write every step clearly, especially in geometry proofs and algebraic derivations.
Use Previous Year Papers Strategically:
Practicing with the 2024 & 2023 question papers is the best way to understand the pattern, difficulty level, and important repeated questions.
Time yourself while solving them to build exam stamina.
Prioritize High-Weightage Chapters: Based on the pattern, give extra focus to:
Chapter 3: Algebra (Graphs, GCD, LCM, Rational Expressions)
Chapter 2: Numbers and Sequences (Euclid's Algorithm, Series)
Chapter 4: Geometry (Pythagoras Theorem, Similarity, Circles)
📥 Essential Resources for Preparation
Official Blueprint & Syllabus: Always refer to the latest syllabus released by the Tamil Nadu State Board to confirm chapters.
Textbook Examples & Exercises: Over 70% of the paper is directly or indirectly based on problems from the state board textbook. Solve every single problem.
Guide Books with Solved Papers: Use reputed guides that provide solved previous year papers and additional practice problems.
Sample Papers & Answer Keys (2023, 2024): These are crucial for practice. While the official 2025 paper will be released after the exam, solving past papers builds confidence.
Official answer key for both tamil and english medium
English
Medium:
1. c –
f∘g = 2/(9x²)
2. c –
91
3. b –
70°
4. b –
2
5. b –
(5, 1)
6. d –
do not intersect
7. d –
11
8. a –
a=3, b=5
9. a –
7
10. d –
x=1, y=-2, z=3
11. b –
5
12. c –
∠B = ∠D
13. a –
8
14. d –
96 cm²
15.
fof(K) = 5, f(K) = 2K–1
f(f(K)) = 2(2K–1)–1 = 4K–3 = 5 → 4K = 8 → K = 2
16.
Poles 6m, 3m, ground AC, find y
By similar triangles: 
→ 6y = 3y + 9 → 3y = 9 → y = 3 m
17.
1 + 2 + … + n = 666
→ 
→ (n–36)(n+37) = 0 → n = 36
18.
Sum to ∞: 9 + 3 + 1 + …
a = 9, r = 1/3 → 
19.
LCM of a²+4a–12, a²–5a+6, GCD = a–2
a²+4a–12 = (a–2)(a+6)
a²–5a+6 = (a–2)(a–3)
LCM = (a–2)(a+6)(a–3)
20.
Father = 6×son; after 6 yrs, father = 4×(son+6)
Let son = x, father = 6x
6x+6 = 4(x+6) → 6x+6 = 4x+24 → 2x = 18 → x=9, 6x=54
Son = 9,
Father = 54
21.
HCF(396, 504, 636) by Euclid
504 – 396 = 108, 396 – 3×108 = 72, 108 – 72 = 36, 72 – 2×36 = 0 →
HCF(396,504)=36
636 ÷ 36 = 17 r 24, 36 ÷ 24 = 1 r 12, 24 ÷ 12 = 2 r 0 → HCF = 12
22.
A.P. 3, 6, 9, …, 111 → terms
a=3, d=3 → 111 = 3 + (n–1)3 → 111 = 3n → n = 37 → 37 terms
23.
R: y = x+3, x∈{0,1,2,3,4,5}
Domain = {0,1,2,3,4,5}
Range = {3,4,5,6,7,8}
24.
LCM: (i) 8x⁴y², 48x³y⁴ → 48x⁴y⁴
(ii)
5x–10 = 5(x–2), 5x²–20 = 5(x–2)(x+2) → LCM = 5(x–2)(x+2)
25.
GCD: 6x³–30x²+60x–48, 3x³–12x²+21x–18
Factor: 6(x–2)³, 3(x–2)²(x–3) → GCD = 3(x–2)²
26.
Sum to ∞: 16, 4, 1, …
a=16, r=1/4 → 
27.
B×A = {(-2,3), (-2,4), (0,3), (0,4), (3,3), (3,4)}
B = { -2, 0, 3 }, A = { 3, 4 }
28.
A=(m,n), B=∅
Tamil Medium:
1.
c) 12
2.
b) (5,1)
3.
a) 7
4.
d) 11
5.
c) 91
6.
b) 5
7.
b) 70°
8.
c)
2/(9x²)
9.
b) 2
10.
d) x=1,
y=-2, z=3
11.
d) do not
intersect / ஒன்றையொன்று வெட்டாது
12.
d) 96 செ.மீ²
13.
c) ∠B = ∠D
14.
a) a=3,
b=5
15.
B×A கொடுக்கப்பட்டுள்ளது → A={3,4}, B={-2,0,3}
16.
fof(K)=5 → f(K)=2K–1 → f(f(K))=4K–3=5 → 4K=8 → K= 2
17.
A.P.
3,6,9,…,111 → a=3,d=3 → 111=3+(n–1)3 →
n= 37
18.
9+3+1+… ∞ → a=9, r=1/3 → S∞ = 9/(1–1/3) = 27/2
19.
தந்தை=6×மகன், 6 ஆண்டுகளுக்குப்
பின் 4 மடங்கு
மகன்=x, தந்தை=6x → 6x+6=4(x+6) → 2x=18 → x=9 → மகன்=9, தந்தை=54
20. A×B=∅,
A×A={(m,m),(m,n),(n,m),(n,n)}
21.
R: y=x+3,
x∈{0,1,2,3,4,5} → மதிப்பகம்={0,1,2,3,4,5}, வீச்சகம்={3,4,5,6,7,8}
22.
1+2+…+n=666 → n(n+1)/2=666 → n²+n–1332=0 →
(n–36)(n+37)=0 → n= 36
23.
a²+4a–12=(a–2)(a+6),
a²–5a+6=(a–2)(a–3) → மீ.சி.ம=(a–2)(a+6)(a–3)
24. உயரங்கள் 6மீ, 3மீ → 6/(y+3)=3/y → 6y=3y+9 → y= 3மீ
25. (i) 8x⁴y², 48x³y⁴ → மீ.சி.ம=48x⁴y⁴
(ii) 5x–10=5(x–2), 5x²–20=5(x–2)(x+2) → மீ.சி.ம=5(x–2)(x+2)
26. மீ.பொ.வ(396,504,636) → 396,504 → மீ.பொ.வ=36 → 36,636 → மீ.பொ.வ= 12
27.
16+4+1+…∞ → a=16, r=1/4 → S∞=16/(1–1/4)= 64/3
28. 6x³–30x²+60x–48, 3x³–12x²+21x–18 → 6(x–2)³, 3(x–2)²(x–3) → மீ.பெ.வ= 3(x–2)²
29. 3+33+333+…n உறுப்புகள்
= 3[1+11+111+…] = 3×(10ⁿ⁺¹–9n–10)/81
30. A={2,3}, B={1}, C={1,2} → B∩C={1}
→ A×(B∩C)={(2,1),(3,1)}
= (A×B)∩(A×C)
31.
வாணி=v, தந்தை=f, தாத்தா=g
(v+f+g)/3=53 → v+f+g=159
g/2+f/3+v/4=65 → 6g+4f+3v=780
g–4=4(v–4) → g=4v–12
தீர்க்க → v= 20,
f= 42, g= 68
32.
4 உறுப்புகள்: a–3d, a–d, a+d, a+3d
கூடுதல்=4a=28 → a=7
வர்க்கங்களின் கூடுதல்=4×49+20d²=276 → d²=4 → d=±2
எண்கள்: 1,5,9,13 அல்லது 13,9,5,1
33. f(x)=x/2–1
f(2)=0, f(4)=1, f(6)=2, f(10)=4, f(12)=5
வரிசைச்சோடிகள்: {(2,0),(4,1),(6,2),(10,4),(12,5)}
34. x³+x²–x+2, 2x³–5x²+5x–3 → மீ.பெ.வ காண → ஈவு 3x²–9x+9
→ மீ.பெ.வ= 1
35. சதுரங்களின் பரப்பு: 10²+11²+…+24²
= [24×25×49/6] – [9×10×19/6] = 4900–285 = 4615 செ.மீ²
36. 3 உறுப்புகள்: a–d, a, a+d
கூடுதல்=3a=27 → a=9
பெருக்கல்=9(81–d²)=288 → d²=49 → d=±7
எண்கள்: 2,9,16 அல்லது 16,9,2
37.
1³+2³+…+8000 → 8000=20³ → கூடுதல்=(20×21/2)²= 44100
38. f(x)=3x–1
f(1)=2, f(2)=5, f(3)=8, f(4)=11
வரிசைச்சோடிகள்: {(1,2),(2,5),(3,8),(4,11)}
39. (i) 67+x≡1(mod4)
→ 67≡3
→ 3+x≡1
→ x≡2
→ x=2
(ii) 8x≡1(mod11) → x=7
40. A={0,1}, B={2,3,4}, C={3,5}
A∪B={0,1,2,3,4} → (A∪B)×C
→ 10 சோடிகள்
(A×C)∪(B×C) → அதே 10 சோடிகள் ✓
41.
(fog)oh =
f(g(h(x))), fo(goh) = f(g(h(x)))
LHS: fog=3x → (fog)oh=3x²
RHS: goh=3x²+1 → f(goh)=3x² ✓
42. 1²+3²+5²+…n உறுப்புகள்
= n(4n²–1)/3
