10th Maths Important Questions: Chapter-wise with Answers (Samacheer Kalvi)

Exam pattern and marks split#

The Maths paper in the Tamil Nadu SSLC public exam is a 100-mark written paper. There is no practical exam in Maths. The pattern in the answer box is the one used in recent public papers, and the March 2026 exam, held on 25 March 2026, followed the same four parts of 1, 2, 5 and 8 marks. The March 2027 SSLC Maths public exam is scheduled for 31 March 2027.

What this split means for your preparation:

• Part III is half the paper. Ten 5-mark answers give 50 marks, so the long sums from every chapter deserve most of your practice time.
• Part II rewards speed and accuracy. Ten 2-mark answers give 20 marks. Most are one-formula or one-step questions, so careless slips cost more than gaps in knowledge.
• Part IV is predictable in type. One question is a geometry construction and the other is a graph. Both carry 8 marks with an internal choice, so you can prepare both types fully.
• Part I needs every chapter. The 14 multiple-choice questions are spread across the book, so do not drop a chapter.

The current Samacheer Kalvi 10th Maths textbook has eight chapters. New textbooks were introduced only for Classes 1 to 3 in 2026-27, and state officials have said revised books for Classes 4 to 12 are planned for 2027-28, so for the 2026-27 year and the March 2027 public exam you study from the existing book.

10th maths chapter wise important questions (state board)#

The questions below are our own textbook-style practice questions, not copies of any guide or past paper. Answers are short; in the exam, write every step, the formula you use and the unit. Take π = 22/7 unless told otherwise.

Chapter 1: Relations and Functions

2 marks

1. If A = {1, 3, 5} and B = {2, 3}, find A × B and B × A. Are they equal? Answer: A × B = {(1,2), (1,3), (3,2), (3,3), (5,2), (5,3)} and B × A = {(2,1), (2,3), (2,5), (3,1), (3,3), (3,5)}. Both have 6 elements, but they are not equal because (1,2) is in A × B and not in B × A.
2. If n(A × B) = 15 and n(A) = 5, find n(B). Answer: 15 ÷ 5 = 3.
3. If f(x) = 2x − 3 and g(x) = x², find f∘g and g∘f. Answer: (f∘g)(x) = 2x² − 3 and (g∘f)(x) = (2x − 3)² = 4x² − 12x + 9, so f∘g ≠ g∘f.

5 marks

4. If f(x) = 3x + 2, g(x) = x − 1 and h(x) = 2x, verify that (f∘g)∘h = f∘(g∘h). Answer: f∘g = 3x − 1, so (f∘g)∘h = 6x − 1. g∘h = 2x − 1, so f∘(g∘h) = 3(2x − 1) + 2 = 6x − 1. Both sides are equal.
5. A linear function f(x) = ax + b has f(1) = 5 and f(3) = 11. Find a, b and f(10). Answer: subtracting gives 2a = 6, so a = 3 and b = 2; f(10) = 32.
6. Let A = {1, 2, 3, 4}, B = {2, 5, 8, 11, 14} and f(x) = 3x − 1. Represent f as an arrow diagram, a set of ordered pairs, a table and a graph. Is f one-one? Is it onto? Answer: f = {(1,2), (2,5), (3,8), (4,11)}. It is one-one (different inputs give different outputs) but not onto, because 14 has no pre-image.

Chapter 2: Numbers and Sequences

2 marks

1. Use Euclid's division algorithm to find the HCF of 336 and 54. Answer: 336 = 54 × 6 + 12; 54 = 12 × 4 + 6; 12 = 6 × 2 + 0, so the HCF is 6.
2. Find the least positive value of x such that 71 ≡ x (mod 8). Answer: 71 = 8 × 8 + 7, so x = 7.
3. Find the 15th term of the AP 3, 7, 11, … Answer: t₁₅ = 3 + 14 × 4 = 59.
4. Find 1³ + 2³ + … + 10³. Answer: (10 × 11 ÷ 2)² = 55² = 3025.

5 marks

5. Find the sum of the first 20 terms of the AP 5, 9, 13, … Answer: S₂₀ = 20/2 × [2 × 5 + 19 × 4] = 10 × 86 = 860.
6. Three numbers in AP have a sum of 27 and a product of 288. Find them. Answer: take a − d, a, a + d. Then 3a = 27, so a = 9, and 9(81 − d²) = 288 gives d² = 49, d = ±7. The numbers are 2, 9, 16.
7. Find the 8th term and the sum of the first 8 terms of the GP 2, 6, 18, … Answer: t₈ = 2 × 3⁷ = 4374; S₈ = 2(3⁸ − 1) ÷ (3 − 1) = 6560.
8. Find the remainder when 2⁸³ is divided by 17. Answer: 2⁴ = 16 ≡ −1 (mod 17), so 2⁸ ≡ 1. Then 2⁸³ = (2⁸)¹⁰ × 2³ ≡ 8. The remainder is 8.

Chapter 3: Algebra

2 marks

1. Find the nature of the roots of 2x² − 4x + 3 = 0. Answer: Δ = b² − 4ac = 16 − 24 = −8 < 0, so there are no real roots.
2. If α and β are the roots of x² − 7x + 10 = 0, find α² + β². Answer: α + β = 7 and αβ = 10, so α² + β² = 49 − 20 = 29.
3. Form the quadratic equation whose roots are 3 and −4. Answer: x² − (sum)x + product = 0 gives x² + x − 12 = 0.
4. Find the GCD and LCM of x² − 1 and x² + 2x + 1. Answer: x² − 1 = (x − 1)(x + 1) and x² + 2x + 1 = (x + 1)², so GCD = x + 1 and LCM = (x − 1)(x + 1)².
5. A has rows (1, 2) and (3, 4); B has rows (0, 1) and (1, 0). Find AB and BA. Answer: AB has rows (2, 1) and (4, 3); BA has rows (3, 4) and (1, 2). So AB ≠ BA.

5 marks

6. Solve x + y + z = 6, x − y + z = 2, 2x + y − z = 1. Answer: subtracting the first two equations gives y = 2, then x + z = 4 and 2x − z = −1, so x = 1 and z = 3.
7. Find the square root of x⁴ − 4x³ + 10x² − 12x + 9. Answer: x² − 2x + 3 (write it with modulus bars, as the textbook does). Check: (x² − 2x + 3)² expands back to the given polynomial.
8. Simplify (x² − 9)/(x² + 5x + 6) ÷ (x − 3)/(x + 2). Answer: (x − 3)(x + 3)/[(x + 2)(x + 3)] × (x + 2)/(x − 3) = 1, for x ≠ −2, −3, 3.

Chapter 4: Geometry

2 marks

1. In triangle ABC, DE ∥ BC with D on AB and E on AC. If AD = 4 cm, DB = 6 cm and AE = 5 cm, find EC. Answer: AD/DB = AE/EC, so 4/6 = 5/EC and EC = 7.5 cm.
2. A man walks 18 m east and then 24 m north. How far is he from the start? Answer: √(18² + 24²) = √900 = 30 m.
3. A point is 13 cm from the centre of a circle of radius 5 cm. Find the length of the tangent from the point. Answer: √(13² − 5²) = √144 = 12 cm.

5 marks

4. State and prove the Basic Proportionality Theorem. Also practise the proofs of the Angle Bisector Theorem, Pythagoras theorem and the Alternate Segment Theorem, and learn the statements of Ceva's and Menelaus' theorems with one application each.
5. AD bisects angle A of triangle ABC. If AB = 6 cm, AC = 9 cm and BD = 4 cm, find DC and BC. Answer: AB/AC = BD/DC, so 6/9 = 4/DC, DC = 6 cm and BC = 10 cm.
6. A student 1.5 m tall casts a 2 m shadow. At the same time a tree casts a 12 m shadow. Find the height of the tree. Answer: the triangles are similar, so 1.5/2 = h/12 and h = 9 m.

Chapter 5: Coordinate Geometry

2 marks

1. Find the slope of the line through (3, −2) and (7, 6). Answer: (6 + 2) ÷ (7 − 3) = 2.
2. Are the lines 2x + 3y = 5 and 3x − 2y = 7 perpendicular? Answer: slopes are −2/3 and 3/2; their product is −1, so yes.
3. Find the equation of the line through (2, 3) with slope 4. Answer: y − 3 = 4(x − 2), that is 4x − y − 5 = 0.

5 marks

4. Find the area of the triangle with vertices (1, 2), (4, 6) and (7, 2). Answer: ½[1(6 − 2) + 4(2 − 2) + 7(2 − 6)] = ½(−24); taking the positive value, the area is 12 sq. units.
5. Find the area of the quadrilateral with vertices (1, 1), (5, 1), (6, 4) and (2, 5). Answer: take the vertices in anticlockwise order and use the shoelace method: ½[(1 + 20 + 30 + 2) − (5 + 6 + 8 + 5)] = ½(53 − 24) = 14.5 sq. units.
6. Show that (2, 3), (4, 5) and (6, 7) are collinear. Answer: area = ½[2(5 − 7) + 4(7 − 3) + 6(3 − 5)] = ½(−4 + 16 − 12) = 0, so the points are collinear.
7. Find the equation of the line through (2, 1) parallel to 3x + y − 5 = 0. Answer: slope −3, so y − 1 = −3(x − 2), that is 3x + y − 7 = 0.

Chapter 6: Trigonometry

2 marks

1. Prove that (1 + tan²θ) cos²θ = 1. Answer: 1 + tan²θ = sec²θ, and sec²θ × cos²θ = 1.
2. A kite string 100 m long makes an angle of 30° with the ground. Find the height of the kite. Answer: 100 × sin 30° = 50 m.

5 marks

3. Prove that sin θ/(1 + cos θ) = (1 − cos θ)/sin θ. Answer: cross-multiplying gives sin²θ = 1 − cos²θ, which is true.
4. Prove that (sec θ − cos θ)(cosec θ − sin θ) = sin θ cos θ. Answer: sec θ − cos θ = sin²θ/cos θ and cosec θ − sin θ = cos²θ/sin θ; their product is sin θ cos θ.
5. From a point 30 m from the foot of a tower, the angle of elevation of its top is 60°. Find the height. Answer: h = 30 tan 60° = 30√3 ≈ 51.96 m.
6. From the top of a 60 m building, the angles of depression of the top and foot of a tower are 30° and 60°. Find the height of the tower. Answer: distance = 60/tan 60° = 20√3 m; the drop to the tower's top = 20√3 × tan 30° = 20 m, so the tower is 60 − 20 = 40 m tall.

Chapter 7: Mensuration

2 marks

1. Find the curved surface area of a cylinder of radius 7 cm and height 10 cm. Answer: 2πrh = 2 × 22/7 × 7 × 10 = 440 cm².
2. Find the total surface area of a solid hemisphere of radius 7 cm. Answer: 3πr² = 3 × 22/7 × 49 = 462 cm².

5 marks

3. Find the volume of a cone of radius 6 cm and height 7 cm. Answer: ⅓ × 22/7 × 36 × 7 = 264 cm³.
4. A metal sphere of radius 6 cm is melted and recast into a cylinder of radius 4 cm. Find the height of the cylinder. Answer: (4/3)π × 216 = π × 16 × h, so h = 18 cm.
5. Find the volume of a frustum with R = 7 cm, r = 3.5 cm and h = 6 cm. Answer: ⅓πh(R² + Rr + r²) = ⅓ × 22/7 × 6 × 85.75 = 539 cm³.
6. A toy is a cone on a hemisphere of the same radius 3.5 cm. The total height is 11.5 cm. Find its volume. Answer: cone height = 11.5 − 3.5 = 8 cm; volume = ⅓πr²h + ⅔πr³ = ⅓πr²(h + 2r) = ⅓ × 22/7 × 12.25 × 15 = 192.5 cm³.

Chapter 8: Statistics and Probability

2 marks

1. Find the range and coefficient of range of 12, 18, 25, 9, 30. Answer: range = 30 − 9 = 21; coefficient = 21 ÷ 39 ≈ 0.54.
2. The mean of a data set is 40 and the standard deviation is 8. Find the coefficient of variation. Answer: 8 ÷ 40 × 100 = 20%.
3. Two dice are thrown. Find the probability that the sum is 7. Answer: 6 favourable outcomes out of 36, so 1/6.

5 marks

4. Find the standard deviation of 2, 4, 6, 8, 10. Answer: mean = 6; squared deviations add up to 16 + 4 + 0 + 4 + 16 = 40; variance = 40 ÷ 5 = 8; SD = √8 = 2√2 ≈ 2.83.
5. A card is drawn from a well-shuffled pack of 52. Find the probability that it is a king or a heart. Answer: 4/52 + 13/52 − 1/52 = 16/52 = 4/13.
6. If P(A) = 0.5, P(B) = 0.3 and P(A ∩ B) = 0.1, find P(A ∪ B) and the probability of neither A nor B. Answer: P(A ∪ B) = 0.5 + 0.3 − 0.1 = 0.7; neither = 1 − 0.7 = 0.3.

Compulsory and repeated questions#

Question 28 in Part II and Question 42 in Part III must be answered; you cannot skip them. They can come from any chapter, which is one more reason not to leave any chapter out. Read them first, because a compulsory question you cannot do affects how you plan the rest of the part.

DGE does not publish a list of "repeated" questions or chapter weightage. What students notice in past papers is that most questions are built on the textbook's worked examples and exercise sums, often with changed numbers. So the most reliable "important questions" list is the textbook itself: every example, every exercise sum and the theorem proofs. Use past public papers (see our guide to the 10th public question paper) to see how those textbook ideas are turned into exam questions.

Geometry and graph questions (Part IV)#

Part IV has two 8-mark questions, each with an either/or choice: Question 43 is a geometry construction and Question 44 is a graph. Prepare every construction and graph type in the textbook so that you can attempt either option.

Constructions to practise: a triangle similar to a given triangle with a scale factor; a triangle given its base, vertical angle and one more measure (median, altitude or angle bisector); tangents from an external point; a tangent using the alternate segment theorem. Write the rough figure, steps of construction and verification.

Graphs to practise:

• Quadratic graph: draw y = x² − x − 6 for x from −3 to 4 (values 6, 0, −4, −6, −6, −4, 0, 6) and read the roots of x² − x − 6 = 0 where the curve cuts the x-axis: x = −2 and x = 3.
• Direct variation: a car moves at a steady 60 km/h. Plot distance against time for 1 to 5 hours; from the straight line read the distance in 3.5 hours (210 km) and the time for 150 km (2.5 hours).
• Inverse variation: 12 workers finish a job in 10 days, so workers × days = 120. Plot the points and read off that 8 workers need 15 days and 15 workers need 8 days.

Choose a sensible scale, label both axes, write the scale in the corner and state the answer in a sentence.

How to practise#

1. Finish the textbook first. Solve every example and exercise sum before buying any 10th samacheer maths guide. A guide is only useful for checking your answer; the method has to be your own.
2. Make a formula sheet for each chapter, especially Mensuration, Coordinate Geometry and Statistics.
3. Do one Part IV question every day in the last two months. Constructions and graphs need neat, unhurried practice.
4. Write timed papers. From January, attempt a full 3-hour paper each week and mark it strictly against the pattern in the answer box.
5. Keep an error log. Note every sign error, wrong formula or skipped step, and revise the log before each test.
6. Balance your languages too. Maths needs daily practice, but do not neglect language papers; our 10th Tamil guide covers the Tamil paper chapter by chapter.
7. Test yourself chapter by chapter. A free account gives you Tamil Nadu State Board chapter-wise lessons and quizzes in English and Tamil medium, so you can check each chapter before moving on.

Planning ahead for Class 12? Our 12th maths important questions follow the same format. For all Tamil Nadu State Board subjects and classes, see the Samacheer Kalvi hub.