10th class mathematics chapter-wise important questions and answers for board exam Pakistan — algebra, trigonometry, geometry, and coordinate geometry solved examples.

10th Class Mathematics Chapter-wise Important Questions and Answers

Quick Answer: The most important 10th class maths chapters for board exams are: Quadratic Equations (often 2-3 questions), Trigonometry (identities and standard angles), Coordinate Geometry (distance, midpoint, slope), and Circle Theorems (proofs and applications). Each chapter has 5-8 question types that repeat regularly across BISE boards.

Knowing which questions matter most is half the battle in maths exam preparation. BISE board papers follow predictable patterns — the same types of questions appear year after year. This chapter-wise guide identifies the most important question types and shows you how to approach them.

Chapter 1: Quadratic Equations — Most Important Questions

Quadratic equations are tested in every board paper and can appear in objective (MCQs), short, and long question sections.

Type 1: Solve Using the Quadratic Formula

Solve: 2x² + 5x − 3 = 0

Solution:

a = 2, b = 5, c = −3
Discriminant: b² − 4ac = 25 − 4(2)(−3) = 25 + 24 = 49
x = (−5 ± √49) / (2×2) = (−5 ± 7) / 4
x = (−5 + 7)/4 = 2/4 = 1/2 OR x = (−5 − 7)/4 = −12/4 = −3
Answer: x = 1/2 or x = −3

Type 2: Find Roots Without Solving (Sum and Product)

If the roots of 3x² − kx + 6 = 0 are equal, find k.

Solution: For equal roots: discriminant = 0 → b² − 4ac = 0 k² − 4(3)(6) = 0 → k² = 72 → k = ±6√2

Type 3: Word Problems Using Quadratic Equations

The product of two consecutive positive integers is 132. Find them.

Solution: Let integers be n and n+1. Then: n(n+1) = 132 n² + n − 132 = 0 (n + 12)(n − 11) = 0 n = 11 (taking positive value) Answer: 11 and 12

Chapter 2: Theory of Quadratic Equations — Common Question Types

Nature of Roots

Find the nature of roots of x² − 6x + 9 = 0

Solution: Discriminant = (−6)² − 4(1)(9) = 36 − 36 = 0 Equal real roots (since discriminant = 0)

Form Equation from Roots

Form the quadratic equation whose roots are 3 and −5.

Solution: Sum of roots = 3 + (−5) = −2 → sum = −b/a so b/a = 2 Product of roots = 3 × (−5) = −15 → product = c/a Equation: x² − (sum)x + (product) = 0 x² − (−2)x + (−15) = 0 x² + 2x − 15 = 0

Chapter 3: Variations — Important Question Types

Direct and Inverse Variation

If y varies directly as x and y = 12 when x = 3, find y when x = 8.

Solution: y/x = constant → 12/3 = y/8 → y = 32

If y varies inversely as x and y = 6 when x = 4, find y when x = 12.

Solution: xy = constant → 6 × 4 = y × 12 → y = 24/12 = 2

Chapter 4: Partial Fractions

Resolve into partial fractions: (2x + 1) / [(x+1)(x−2)]

Solution: (2x + 1) / [(x+1)(x−2)] = A/(x+1) + B/(x−2)

Multiply both sides by (x+1)(x−2): 2x + 1 = A(x−2) + B(x+1)

Put x = 2: 5 = 3B → B = 5/3 Put x = −1: −1 = −3A → A = 1/3

Answer: (1/3)/(x+1) + (5/3)/(x−2)

Chapter 5: Sets and Functions

Important Set Questions

If A = and B = , find A∪B, A∩B, A−B, B−A.

Solution:

A∪B = (all elements from both)
A∩B = (common elements)
A−B = (in A but not B)
B−A = (in B but not A)

Verify: n(A∪B) = n(A) + n(B) − n(A∩B) = 5 + 5 − 3 = 7 ✓

Chapter 6: Trigonometric Identities — Common Board Questions

Type 1: Prove the Identity

Prove: sin²θ / (1 − cos θ) = 1 + cos θ

Solution (working on LHS): LHS = sin²θ / (1 − cos θ) = (1 − cos²θ) / (1 − cos θ) [using sin²θ = 1 − cos²θ] = (1 − cos θ)(1 + cos θ) / (1 − cos θ) [factoring numerator] = 1 + cos θ = RHS ✓

Type 2: Find Value Using Given Information

If sin A = 3/5, find cos A and tan A (A is in first quadrant).

Solution: sin²A + cos²A = 1 (3/5)² + cos²A = 1 cos²A = 1 − 9/25 = 16/25 cos A = 4/5 (positive, first quadrant) tan A = sin A / cos A = (3/5)/(4/5) = 3/4

Chapter 7: Coordinate Geometry — Must-Know Questions

Finding Equation of a Line

Find the equation of the line passing through (2, 3) and (4, 7).

Solution: Slope m = (7 − 3)/(4 − 2) = 4/2 = 2 Using point-slope form: y − 3 = 2(x − 2) y − 3 = 2x − 4 y = 2x − 1

Distance and Midpoint

Find the distance between A(1, 2) and B(4, 6).

Solution: d = √((4−1)² + (6−2)²) = √(9 + 16) = √25 = 5 units

Midpoint of AB = ((1+4)/2, (2+6)/2) = (2.5, 4)

Chapter 8: Circle Theorems — Proving Questions

Circle theorems are long questions in board exams. You must know the statement and proof.

Theorem: Angle at Centre = 2 × Angle at Circumference

Statement: The central angle is double any inscribed angle that subtends the same arc.

Given: Circle with centre O; arc AB; angle AOB at centre; angle ACB at circumference (C on major arc).

To prove: Angle AOB = 2 × Angle ACB

Proof Outline:

Draw line OC extended to point D
Triangle OAC is isosceles (OA = OC = radii), so angle OAC = angle OCA
Exterior angle AOD = angle OAC + angle OCA = 2 × angle OCA
Similarly, angle BOD = 2 × angle OCB
angle AOB = angle AOD + angle BOD = 2(angle OCA + angle OCB) = 2 × angle ACB ✓

For complete marks in geometry proofs, present the proof in clearly numbered steps with reasons given for each statement.

Study Strategy for Chapter-wise Preparation

Weeks 4-3 before exam: Cover each chapter systematically. Solve all exercise questions in textbook.

Weeks 2-1 before exam: Focus on past paper questions from each chapter. Identify which question types appear most often.

Final week: Practice complete papers under timed conditions. Identify weak areas and address them.

Day before exam: Review key formulas and theorem statements only. Do not solve new problems.

Check your board exam results and see how your maths performance affects your overall grade at /tools/gpa-calculator and /10th-class-result-2026.

Frequently Asked Questions

Q: Which chapter of 10th class maths carries the most marks in board exams? A: Trigonometry and coordinate geometry together typically contribute the most to Section C (long questions). Quadratic equations appear across all sections. Prioritize these three chapters.

Q: How do I prepare for maths MCQs in the board exam? A: MCQs in maths test formula knowledge and quick calculation. Practice each formula with number substitution. Past paper MCQs are the best resource — patterns repeat.

Q: Should I attempt all questions or skip hard ones? A: In Section B (short questions), you usually have a choice (e.g., attempt 6 out of 9). In Section C (long questions), you usually choose 3 from 5. Always attempt questions you are most confident about first.

Q: How much time should I spend on each section during the exam? A: A typical allocation: 15 minutes for Section A (MCQs), 45 minutes for Section B, 45 minutes for Section C, 15 minutes for review. Adjust based on your speed, but do not spend more than 12 minutes on any single long question.

Q: What if I am stuck on a numerical in the exam? A: Write what you know — the given information, the formula, and as many steps as you can. Partial marks are better than leaving a blank. Then move on and come back if time allows.

10th Class Mathematics Chapter-wise Important Questions and Answers

Chapter 1: Quadratic Equations — Most Important Questions

Quadratic equations are tested in every board paper and can appear in objective (MCQs), short, and long question sections.

Type 1: Solve Using the Quadratic Formula

Solve: 2x² + 5x − 3 = 0

Solution:

a = 2, b = 5, c = −3
Discriminant: b² − 4ac = 25 − 4(2)(−3) = 25 + 24 = 49
x = (−5 ± √49) / (2×2) = (−5 ± 7) / 4
x = (−5 + 7)/4 = 2/4 = 1/2 OR x = (−5 − 7)/4 = −12/4 = −3
Answer: x = 1/2 or x = −3

Type 2: Find Roots Without Solving (Sum and Product)

If the roots of 3x² − kx + 6 = 0 are equal, find k.

Solution: For equal roots: discriminant = 0 → b² − 4ac = 0 k² − 4(3)(6) = 0 → k² = 72 → k = ±6√2

Type 3: Word Problems Using Quadratic Equations

The product of two consecutive positive integers is 132. Find them.

Solution: Let integers be n and n+1. Then: n(n+1) = 132 n² + n − 132 = 0 (n + 12)(n − 11) = 0 n = 11 (taking positive value) Answer: 11 and 12

Chapter 2: Theory of Quadratic Equations — Common Question Types

Nature of Roots

Find the nature of roots of x² − 6x + 9 = 0

Solution: Discriminant = (−6)² − 4(1)(9) = 36 − 36 = 0 Equal real roots (since discriminant = 0)

Form Equation from Roots

Form the quadratic equation whose roots are 3 and −5.

Chapter 3: Variations — Important Question Types

Direct and Inverse Variation

If y varies directly as x and y = 12 when x = 3, find y when x = 8.

Solution: y/x = constant → 12/3 = y/8 → y = 32

If y varies inversely as x and y = 6 when x = 4, find y when x = 12.

Solution: xy = constant → 6 × 4 = y × 12 → y = 24/12 = 2

Chapter 4: Partial Fractions

Resolve into partial fractions: (2x + 1) / [(x+1)(x−2)]

Solution: (2x + 1) / [(x+1)(x−2)] = A/(x+1) + B/(x−2)

Multiply both sides by (x+1)(x−2): 2x + 1 = A(x−2) + B(x+1)

Put x = 2: 5 = 3B → B = 5/3 Put x = −1: −1 = −3A → A = 1/3

Answer: (1/3)/(x+1) + (5/3)/(x−2)

Chapter 5: Sets and Functions

Important Set Questions

If A = and B = , find A∪B, A∩B, A−B, B−A.

Solution:

A∪B = (all elements from both)
A∩B = (common elements)
A−B = (in A but not B)
B−A = (in B but not A)

Verify: n(A∪B) = n(A) + n(B) − n(A∩B) = 5 + 5 − 3 = 7 ✓

Chapter 6: Trigonometric Identities — Common Board Questions

Type 1: Prove the Identity

Prove: sin²θ / (1 − cos θ) = 1 + cos θ

Type 2: Find Value Using Given Information

If sin A = 3/5, find cos A and tan A (A is in first quadrant).

Solution: sin²A + cos²A = 1 (3/5)² + cos²A = 1 cos²A = 1 − 9/25 = 16/25 cos A = 4/5 (positive, first quadrant) tan A = sin A / cos A = (3/5)/(4/5) = 3/4

Chapter 7: Coordinate Geometry — Must-Know Questions

Finding Equation of a Line

Find the equation of the line passing through (2, 3) and (4, 7).

Solution: Slope m = (7 − 3)/(4 − 2) = 4/2 = 2 Using point-slope form: y − 3 = 2(x − 2) y − 3 = 2x − 4 y = 2x − 1

Distance and Midpoint

Find the distance between A(1, 2) and B(4, 6).

Solution: d = √((4−1)² + (6−2)²) = √(9 + 16) = √25 = 5 units

Midpoint of AB = ((1+4)/2, (2+6)/2) = (2.5, 4)

Chapter 8: Circle Theorems — Proving Questions

Circle theorems are long questions in board exams. You must know the statement and proof.

Theorem: Angle at Centre = 2 × Angle at Circumference

Statement: The central angle is double any inscribed angle that subtends the same arc.

Given: Circle with centre O; arc AB; angle AOB at centre; angle ACB at circumference (C on major arc).

To prove: Angle AOB = 2 × Angle ACB

Proof Outline:

Draw line OC extended to point D
Triangle OAC is isosceles (OA = OC = radii), so angle OAC = angle OCA
Exterior angle AOD = angle OAC + angle OCA = 2 × angle OCA
Similarly, angle BOD = 2 × angle OCB
angle AOB = angle AOD + angle BOD = 2(angle OCA + angle OCB) = 2 × angle ACB ✓

For complete marks in geometry proofs, present the proof in clearly numbered steps with reasons given for each statement.

Study Strategy for Chapter-wise Preparation

Weeks 4-3 before exam: Cover each chapter systematically. Solve all exercise questions in textbook.

Weeks 2-1 before exam: Focus on past paper questions from each chapter. Identify which question types appear most often.

Final week: Practice complete papers under timed conditions. Identify weak areas and address them.

Day before exam: Review key formulas and theorem statements only. Do not solve new problems.

Check your board exam results and see how your maths performance affects your overall grade at /tools/gpa-calculator and /10th-class-result-2026.