CBSE Class 10 Mathematics: All Chapters Important Questions Maths PDF (2025-2026)

Mastering CBSE Class 10 Mathematics for the 2025–2026 board exam requires consistent, targeted problem-solving across all 14 NCERT chapters. Preparing with a structured CBSE Class 10 All Chapters Important Questions Maths PDF blueprint empowers students to identify recurring question types, master NCERT exemplar challenges, and secure maximum marks through step-by-step solutions aligned with the official CBSE marking scheme.

Chapter-Wise Key Concepts & Formula Blueprint

The Class 10 Mathematics curriculum is structured into seven fundamental units. Achieving a score of 95+ requires absolute clarity on definitions, theorem proofs, and standard algebraic formulas.

1. Number Systems (Real Numbers)

The unit centers on the Fundamental Theorem of Arithmetic, which states that every composite number can be expressed uniquely as a product of primes, apart from the order in which the prime factors occur.

• Fundamental Relationship: For any two positive integers a and b, HCF(a, b) × LCM(a, b) = a × b.
• Irrationality Proofs: Rigorous contradiction proofs showing that numbers like √2, √3, √5, and 3 + 2√5 are irrational.

2. Algebra (Polynomials, Linear Equations, Quadratic Equations, AP)

Algebra carries the highest weightage (20 marks) in the board exam 10. Key relationships include:

• Quadratic Polynomials: For p(x) = ax² + bx + c with zeroes α and β:
  • Sum of zeroes: α + β = -b/a
  • Product of zeroes: αβ = c/a
• Pair of Linear Equations: Consistency conditions for a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0:
  • Unique solution (intersecting lines): a1/a2 ≠ b1/b2
  • Infinitely many solutions (coincident lines): a1/a2 = b1/b2 = c1/c2
  • No solution (parallel lines/inconsistent): a1/a2 = b1/b2 ≠ c1/c2
• Quadratic Equations: Roots of ax² + bx + c = 0 are given by the quadratic formula:
  x = (-b ± √(b2 - 4ac)) / (2a)
  The discriminant D = b2 - 4ac determines root characteristics:
  • D > 0: Two distinct real roots
  • D = 0: Two equal real roots (x = -b/(2a))
  • D < 0: No real roots
• Arithmetic Progressions (AP):
  • n^th term: an = a + (n - 1)d
  • Sum of first n terms: Sn = (n/2)[2a + (n - 1)d] = (n/2)[a + l]

3. Coordinate Geometry

• Distance Formula: d = √[(x2 - x1)2 + (y2 - y1)2]
• Section Formula (Internal Division): P(x, y) = ((m1x2 + m2x1)/(m1 + m2), (m1y2 + m2y1)/(m1 + m2))
• Midpoint Formula: M = ((x1 + x2)/2, (y1 + y2)/2)

4. Geometry (Triangles & Circles)

• Basic Proportionality Theorem (Thales' Theorem): If a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio.
• Tangent Properties:
  • The tangent at any point of a circle is perpendicular to the radius through the point of contact.
  • The lengths of tangents drawn from an external point to a circle are equal.

5. Trigonometry & Applications

• Standard Identities:
  • sin2θ + cos2θ = 1
  • 1 + tan2θ = sec2θ
  • 1 + cot2θ = cosec2θ
• Heights and Distances: Resolving right-angled triangles using angle of elevation and angle of depression.

6. Mensuration (Surface Areas and Volumes & Areas Related to Circles)

• Sector Area: A = (θ/360°) × πr2; Arc Length: l = (θ/360°) × 2πr
• Combination of Solids: Total surface area and volume of combined cylinders, cones, spheres, and hemispheres.

7. Statistics and Probability

• Mean (μ): Direct Method: Σ(fixi) / Σfi; Assumed Mean Method: a + Σ(fidi) / Σfi
• Median: L + [((N/2) - cf) / f] × h
• Mode: L + [(f1 - f0) / (2f1 - f0 - f2)] × h
• Empirical Relationship: 3 × Median = Mode + 2 × Mean
• Theoretical Probability: P(E) = (Number of favorable outcomes) / (Total number of possible outcomes), where 0 ≤ P(E) ≤ 1.

Important CBSE Questions with Step-by-Step Solutions

The following problems represent frequently tested patterns across standard and basic CBSE Class 10 Mathematics board examinations.

Question 1: Real Numbers (Proof of Irrationality — 3 Marks)

Problem: Prove that √5 is an irrational number.

Solution:

1. Let us assume to the contrary that √5 is rational.
2. Therefore, we can find co-prime integers a and b (where b ≠ 0) such that:
   √5 = a / b ⇒ a = b√5
3. Squaring both sides:
   a2 = 5b2 … (Equation 1)
   This implies that 5 divides a². By the fundamental theorem of arithmetic, if a prime p divides a², then p divides a. Hence, 5 divides a.
4. Let a = 5c for some integer c. Substituting in Equation 1:
   (5c)2 = 5b2 ⇒ 25c2 = 5b2 ⇒ b2 = 5c2
5. This implies that 5 divides b², which means 5 divides b.
6. From steps 3 and 5, 5 is a common factor of both a and b. This contradicts the fact that a and b are co-prime (having no common factor other than 1).
7. This contradiction has arisen because of our incorrect assumption that √5 is rational. Hence, √5 is irrational. (Proved)

Question 2: Arithmetic Progression (Real-World Application — 4 Marks)

Problem: The sum of the 4th and 8th terms of an AP is 24, and the sum of the 6th and 10th terms is 44. Find the first three terms and the 20th term of the AP.

Solution:

1. Let the first term be a and the common difference be d.
   Recall that an = a + (n - 1)d.
2. Given: a4 + a8 = 24
(a + 3d) + (a + 7d) = 24
2a + 10d = 24 ⇒ a + 5d = 12 … (Equation 1)
3. Given: a6 + a10 = 44
(a + 5d) + (a + 9d) = 44
2a + 14d = 44 ⇒ a + 7d = 22 … (Equation 2)
4. Subtracting Equation 1 from Equation 2:
   (a + 7d) - (a + 5d) = 22 - 12
2d = 10 ⇒ d = 5
5. Substitute d = 5 into Equation 1:
   a + 5(5) = 12 ⇒ a + 25 = 12 ⇒ a = -13
6. The first three terms are:
   • a₁ = -13
   • a₂ = -13 + 5 = -8
   • a₃ = -8 + 5 = -3
7. Finding the 20th term:
   a20 = a + 19d = -13 + 19(5) = -13 + 95 = 82.

Answer: First three terms are -13, -8, -3 and the 20th term is 82.

Question 3: Trigonometric Identity (Analytical Proof — 4 Marks)

Problem: Prove the identity: (sin θ - cos θ + 1) / (sin θ + cos θ - 1) = 1 / (sec θ - tan θ) using the identity sec2θ = 1 + tan2θ.

Solution:

1. Divide numerator and denominator of LHS by cos θ:
   LHS = [(sin θ/cos θ) - (cos θ/cos θ) + (1/cos θ)] / [(sin θ/cos θ) + (cos θ/cos θ) - (1/cos θ)]
LHS = (tan θ - 1 + sec θ) / (tan θ + 1 - sec θ) = (tan θ + sec θ - 1) / (tan θ - sec θ + 1)
2. Using 1 = sec2θ - tan2θ = (sec θ - tan θ)(sec θ + tan θ) in the numerator:
   Numerator = (sec θ + tan θ) - (sec θ - tan θ)(sec θ + tan θ)
= (sec θ + tan θ)[1 - (sec θ - tan θ)] = (sec θ + tan θ)(1 - sec θ + tan θ)
3. Now substitute back:
   LHS = [(sec θ + tan θ)(tan θ - sec θ + 1)] / (tan θ - sec θ + 1)
   Cancelling the common factor (tan θ - sec θ + 1):
   LHS = sec θ + tan θ
4. Multiply and divide by (sec θ - tan θ):
   LHS = [(sec θ + tan θ)(sec θ - tan θ)] / (sec θ - tan θ) = (sec2θ - tan2θ) / (sec θ - tan θ)
   Since sec2θ - tan2θ = 1:
   LHS = 1 / (sec θ - tan θ) = RHS. (Proved)

Question 4: Surface Areas and Volumes (5 Marks)

Problem: A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 4 cm and the diameter of the base is 8 cm. Determine the volume of the toy. If a cube of edge 10 cm circumscribes the toy, find the difference between the volume of the cube and the toy. (Take π = 3.14)

Solution:

1. Dimensions:
   Radius of cone and hemisphere (r) = 8 / 2 = 4 cm.
   Height of cone (h) = 4 cm.
2. Volume of the toy (V_toy):
   Vtoy = Volume of Cone + Volume of Hemisphere
Vtoy = (1/3)πr2h + (2/3)πr3 = (1/3)πr2(h + 2r)
Vtoy = (1/3) × 3.14 × (4)2 × (4 + 2 × 4)
Vtoy = (1/3) × 3.14 × 16 × 12 = 3.14 × 16 × 4 = 200.96 cm3
3. Volume of the cube (V_cube):
   Vcube = side3 = 103 = 1000 cm3
4. Difference in volume:
   Vdiff = 1000 - 200.96 = 799.04 cm3

Answer: Volume of the toy is 200.96 cm³ and the difference in volume is 799.04 cm³.

Question 5: Statistics (Missing Frequency — 4 Marks)

Problem: The median of the following distribution is 525. Find the values of x and y if the total frequency is 100.

Solution:

1. Total frequency N = 100.
   76 + x + y = 100 ⇒ x + y = 24 … (Equation 1)
2. Given Median = 525, which lies in the class interval 500 – 600.
   Lower limit of median class (L) = 500
   Class size (h) = 100
   Frequency of median class (f) = 20
   Cumulative frequency of preceding class (cf) = 36 + x
   N / 2 = 50
3. Using the median formula:
   Median = L + [((N/2) - cf) / f] × h
525 = 500 + [(50 - (36 + x)) / 20] × 100
25 = (14 - x) × 5
5 = 14 - x ⇒ x = 9
4. Substituting x = 9 into Equation 1:
   9 + y = 24 ⇒ y = 15

Answer: x = 9 and y = 15.

Strategic Preparation Advice for CBSE Class 10 Board Exams

A structured study system separates top scorers from average performers. Implement these subject-specific strategies during your preparation:

1. Master NCERT Examples First: Over 40% of standard questions in board exam 10 papers are drawn directly from NCERT solved examples or miscellaneous exercises.
2. Maintain a Step-Marking Habit: CBSE examiners award marks for:
   • Writing the standard formula correctly (0.5 to 1 mark).
   • Clear substitution of values with units.
   • Concluding with a distinct final statement highlighting the answer in a box.
3. Master Theorems with Diagrams: Practice drawing neat, labelled geometric figures with a ruler and pencil for the Basic Proportionality Theorem and Tangents from an External Point. A missing diagram often leads to penalty deductions.
4. Tackle Case-Study Questions: Case-based questions (12 marks total) test real-life mathematical modelling in Coordinate Geometry, Applications of Trigonometry, and Arithmetic Progressions. Break these 4-mark questions into sub-parts systematically.

Where to Practice More & Download Question Papers

Solving authentic sample papers under timed 3-hour examination conditions is the single most effective way to eliminate calculation errors and master time management. Practice with chapter-wise tests, assertion-reasoning worksheets, and full-length CBSE mock papers complete with detailed step-wise marking rubrics directly at Theorify QPTool (qptool.theorify.in).