NCERT Math Solution Class 10 | Download Free PDF

Class 10 Maths is the first time students sit a subject in a high-stakes board examination format, and the Class 10 Maths Syllabus demands more than calculation ability. Proof writing, graph construction, and formula application all carry independent marks. These NCERT Solutions for Class 10 Maths cover all chapters of the current CBSE pattern with step-by-step answers written to match the exact format and method CBSE expects in exams, including partial marking for correct steps.

NCERT Solutions for Class 10 Maths - Chapter-wise PDF

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Some sub-topics within Chapter 3 (Pair of Linear Equations) and Chapter 11 (Areas Related to Circles) have been removed from the NCERT textbook but remain in the CBSE Class 10 2026-27 Syllabus. These are covered in the solutions.

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Class 10 Maths - Unit-wise Weightage

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NOTE: Algebra is the heaviest unit. Triangles and Circles together carry proof-based questions worth 5 marks each that appear without exception in every board paper. Trigonometry is highly predictable, as the identity-proof questions follow a narrow set of patterns. Statistics and Probability are straightforward and should not be left for last minute.

Important topics Covered in NCERT Solutions Class 10

Every chapter solution includes answers to all NCERT exercises, worked examples, and additional questions where applicable. Check the Class 10 Maths Important Questions to see which question types appear most frequently. Here is what each chapter covers and what to focus on. 

Chapter 1: Real Numbers

Key topics: Euclid's Division Algorithm and HCF, Fundamental Theorem of Arithmetic, prime factorisation, LCM and HCF using prime factors, proofs of irrationality for numbers like √2, √3 and √5, terminating and non-terminating recurring decimals.

Example Question: Prove that √3 is irrational.

Solution: Assume √3 is rational. Then √3 = p/q, where p and q are integers, q ≠ 0, and p and q are coprime (HCF = 1).

• Squaring both sides: 3 = p²/q², so p² = 3q².
• This means 3 divides p². Since 3 is prime, 3 divides p. Let p = 3c.
• Substituting: (3c)² = 3q², so 9c² = 3q², so q² = 3c².
• This means 3 divides q². Since 3 is prime, 3 divides q.
• 3 divides both p and q, which contradicts the assumption that p and q are coprime. Therefore √3 is irrational.

Chapter 2: Polynomials

Key topics: Zeroes of a polynomial and their geometric meaning through graph interpretation, relationship between zeroes and coefficients for quadratic polynomials (sum = -b/a, product = c/a) and cubic polynomials, division algorithm.

Example Question: If α and β are the zeroes of the polynomial 2x² + 5x + k, and α² + β² + αβ = 21/4, find the value of k.

Solution: For 2x² + 5x + k:

• α + β = -5/2 and αβ = k/2.
• α² + β² = (α + β)² - 2αβ = 25/4 - k.
• So α² + β² + αβ = 25/4 - k + k/2 = 25/4 - k/2.
• Setting equal to 21/4: 25/4 - k/2 = 21/4.
• -k/2 = -1, therefore k = 2.

Chapter 3: Pair of Linear Equations in Two Variables

Key topics: Graphical method with intersecting lines (unique solution), parallel lines (no solution), and coincident lines (infinitely many solutions). Algebraic methods: substitution, elimination, cross-multiplication. Consistency conditions using ratios a1/a2, b1/b2, c1/c2. Word problems requiring formation of two equations from text.

Example Question: Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. Find their present ages.

Solution: Let the present age of Nuri = x years and Sonu = y years.

• Five years ago: x - 5 = 3(y - 5), which gives x - 3y = -10 ... (1).
• Ten years later: x + 10 = 2(y + 10), which gives x - 2y = 10 ... (2).
• Subtracting (1) from (2): y = 20.
• Substituting y = 20 in (2): x - 40 = 10, so x = 50.
• Nuri's present age is 50 years and Sonu's present age is 20 years.

Chapter 4: Quadratic Equations

Key topics: Standard form ax² + bx + c = 0. Factorisation method. Completing the square. Quadratic formula x = (-b ± √(b²-4ac)) / 2a. Discriminant (b²-4ac) and nature of roots: two distinct real roots when D > 0, two equal roots when D = 0, no real roots when D < 0. Word problems translating into quadratic equations.

Example Question: A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Solution: Let the speed of the train = x km/h.

• Time at speed x = 360/x hours. Time at speed (x + 5) = 360/(x + 5) hours.
• Given: 360/x - 360/(x + 5) = 1.
• 360(x + 5) - 360x = x(x + 5), so 1800 = x² + 5x.
• x² + 5x - 1800 = 0.
• Factorising: (x - 40)(x + 45) = 0, so x = 40 or x = -45.
• Since speed cannot be negative, x = 40. The speed of the train is 40 km/h.

Chapter 5: Arithmetic Progressions

Key topics: General term: an = a + (n-1)d. Sum of n terms: Sn = n/2 x (2a + (n-1)d). Finding first term and common difference from partial information. Problems where both formulae must be applied together to find an unknown.

Example Question: The sum of the first 7 terms of an AP is 63 and the sum of its next 7 terms is 161. Find the AP.

Solution: Let the first term = a and common difference = d.

• S7 = 7/2 x (2a + 6d) = 63, which gives a + 3d = 9 ... (1).
• Sum of first 14 terms = 63 + 161 = 224, so 14/2 x (2a + 13d) = 224, giving 2a + 13d = 32 ... (2).
• Multiplying (1) by 2: 2a + 6d = 18 ... (3).
• Subtracting (3) from (2): 7d = 14, so d = 2.
• Substituting in (1): a + 6 = 9, so a = 3.
• The AP is 3, 5, 7, 9, 11, ...

Chapter 6: Triangles

Key topics: Basic Proportionality Theorem (Thales' theorem), statement, proof and converse. Criteria for similarity: AA, SAS, SSS. Pythagoras theorem, statement, proof and converse. Ratio of areas of similar triangles (ratio of squares of corresponding sides).

Example Question: In triangle ABC, DE is parallel to BC. If AD = 1.5 cm, DB = 3 cm, and AE = 1 cm, find EC.

Solution: Since DE is parallel to BC, by the Basic Proportionality Theorem:

• AD/DB = AE/EC.
• Substituting: 1.5/3 = 1/EC.
• EC = 3 x 1/1.5 = 2 cm.

Chapter 7: Coordinate Geometry

Key topics: Distance formula: d = √((x2-x1)² + (y2-y1)²). Section formula for internal division: ((m1x2 + m2x1)/(m1+m2), (m1y2 + m2y1)/(m1+m2)). Midpoint formula as a special case. Area of a triangle using coordinates: 1/2 |x1(y2-y3) + x2(y3-y1) + x3(y1-y2)|. Collinearity condition when area equals 0.

Example Question: Find the ratio in which the point P(x, 2) divides the line segment joining A(-2, -2) and B(3, 7). Also find the value of x.

Solution: Let P divide AB in the ratio k:1.

• Using section formula for y-coordinate: (7k - 2)/(k + 1) = 2.
• 7k - 2 = 2k + 2, so 5k = 4, giving k = 4/5.
• The ratio is 4:5.
• Using section formula for x-coordinate: x = (4/5 x 3 + (-2)) / (4/5 + 1).
• x = (12/5 - 10/5) / (9/5) = (2/5) / (9/5) = 2/9.

Chapter 8: Introduction to Trigonometry

Key topics: Trigonometric ratios sin, cos, tan, cosec, sec and cot defined for acute angles of a right triangle. Values for standard angles 0°, 30°, 45°, 60° and 90°. Ratios of complementary angles: sin θ = cos(90°-θ) and related pairs. Three Pythagorean identities: sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, 1 + cot²θ = cosec²θ.

Example Question: Prove that (sin θ + cosec θ)² + (cos θ + sec θ)² = 7 + tan²θ + cot²θ.

Solution: Starting with the LHS:

• (sin θ + cosec θ)² + (cos θ + sec θ)²
• = sin²θ + 2 sin θ cosec θ + cosec²θ + cos²θ + 2 cos θ sec θ + sec²θ.
• Since sin θ cosec θ = 1 and cos θ sec θ = 1: = (sin²θ + cos²θ) + 4 + cosec²θ + sec²θ.
• = 1 + 4 + (1 + cot²θ) + (1 + tan²θ).
• = 7 + tan²θ + cot²θ = RHS.

Chapter 9: Some Applications of Trigonometry

Key topics: Angle of elevation. Angle of depression. Line of sight. Single-triangle problems finding the height of a tower or distance from a point. Two-triangle problems where the observer moves, two observations are made from different positions of the same object, and the unknown is solved across both triangles.

Example Question: The angles of elevation of the top of a tower from two points at distances a and b from the base, on the same side, are 60° and 30° respectively. Prove that the height of the tower is √(ab).

Solution: Let the height of the tower = h.

• From the point at distance b (angle 30°): tan 30° = h/b, so 1/√3 = h/b, giving b = h√3 ... (1).
• From the point at distance a (angle 60°): tan 60° = h/a, so √3 = h/a, giving a = h/√3 ... (2).
• Multiplying (1) and (2): ab = h√3 x h/√3 = h².
• Therefore h = √(ab). Hence proved.

Chapter 10: Circles

Key topics: Tangent to a circle, definition and point of tangency. Theorem 1: the tangent at any point on a circle is perpendicular to the radius at that point. Theorem 2: tangents drawn from an external point to a circle are equal in length, with proof. Number of tangents from a point on, inside, and outside the circle.

Example Question: Two tangents PA and PB are drawn from an external point P to a circle with centre O. Prove that PA = PB.

Solution: In triangles OAP and OBP:

• OA = OB (radii of the same circle).
• OP = OP (common side).
• Angle OAP = Angle OBP = 90° (radius is perpendicular to tangent at point of contact).
• By RHS congruence criterion, triangle OAP is congruent to triangle OBP.
• Therefore PA = PB (corresponding parts of congruent triangles).

Chapter 11: Areas Related to Circles

Key topics: Length of an arc: (θ/360°) x 2πr. Area of a sector: (θ/360°) x πr². Area of a minor segment = area of sector minus area of triangle. Area of combination figures such as a circle inscribed in a square, ring or annulus, flower-petal shapes, and semicircle with rectangle or triangle.

Example Question: A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding minor segment. (Use π = 3.14)

Solution: Radius r = 10 cm, angle θ = 90°.

• Area of sector = (90/360) x 3.14 x 10 x 10 = 1/4 x 314 = 78.5 cm².
• Area of triangle formed by the two radii and the chord = 1/2 x 10 x 10 = 50 cm².
• Area of minor segment = 78.5 - 50 = 28.5 cm².

Chapter 12: Surface Areas and Volumes

Key topics: Curved surface area and total surface area of cylinder, cone, sphere and hemisphere. Volume of each. Combined solids such as cone mounted on cylinder, hemisphere mounted on cylinder, cone inscribed in sphere, and hemisphere embedded in cuboid. Conversion problems where a solid is melted and recast into a different shape or into multiple smaller solids.

Example Question: A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm³ of iron has approximately 8 g mass. (Use π = 3.14)

Solution:

• Volume of larger cylinder (r = 12 cm, h = 220 cm): V1 = 3.14 x 12 x 12 x 220 = 99475.2 cm³.
• Volume of smaller cylinder (r = 8 cm, h = 60 cm): V2 = 3.14 x 8 x 8 x 60 = 12057.6 cm³.
• Total volume = 99475.2 + 12057.6 = 111532.8 cm³.
• Mass = 111532.8 x 8 = 892262.4 g = 892.26 kg.

Chapter 13: Statistics

Key topics: Mean of grouped data using direct method, assumed mean method and step deviation method. Median of grouped data: identifying the median class using cumulative frequency, interpolation formula Median = l + ((n/2 - cf)/f) x h. Mode of grouped data: modal class formula Mode = l + ((f1 - f0)/(2f1 - f0 - f2)) x h. Cumulative frequency of less than type and more than type. Ogive construction and graphical determination of median from the intersection of both ogives.

Example Question: Find the mean of the following distribution using the assumed mean method.

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Solution: Taking assumed mean A = 25.

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Mean = A + (Σfi x di / Σfi) = 25 + (140/50) = 25 + 2.8 = 27.8.

Chapter 14: Probability

Key topics: Classical probability: P(E) = (number of favourable outcomes) / (total number of equally likely outcomes). Complementary events: P(E) + P(Ē) = 1. Impossible events where P = 0 and certain events where P = 1. Problems based on a standard deck of 52 playing cards with 4 suits and 13 cards each, including face cards and aces. Problems based on coins, dice, and numbered cards or balls.

Example Question: A card is drawn at random from a well-shuffled deck of 52 playing cards. Find the probability that the card drawn is (i) a king, (ii) not a king, (iii) a red ace.

Solution: Total number of cards = 52.

• (i) Number of kings = 4, so P(king) = 4/52 = 1/13.
• (ii) P(not a king) = 1 - 1/13 = 12/13.
• (iii) Number of red aces = 2 (ace of hearts and ace of diamonds), so P(red ace) = 2/52 = 1/26.

How to Score Well in Class 10 Maths Exams?

The Maths paper is 80 marks and CBSE awards marks for steps, not just answers. A wrong final answer with correct working still earns partial marks. A correct answer with no working shown earns nothing in most question types.

1. For proof questions in Triangles and Circles: state the theorem or property being used at every step. Every step must cite its justification. Do not assume the examiner will infer the reasoning.
2. For numerical questions in Mensuration, Statistics and AP: write the formula, then substitute values with units, then calculate. Never skip to the answer directly. Each stage is a potential mark.
3. For graph questions in Linear Equations, Coordinate Geometry and Statistics: label axes, mark coordinates clearly, and draw a neat line or curve. Scale must be consistent and stated. Presentation directly affects marks in these questions.
4. For identity proofs in Trigonometry: choose which side to simplify and work toward the other. Working on both sides simultaneously is not accepted as valid proof in board marking.
5. For word problems in Quadratic Equations, AP and Coordinate Geometry: write the equation or expression being formed before solving. Marks are allocated for correct formation separately from correct solutions.

The exam pattern is consistent year on year. Going through Class 10 Previous Year Papers alongside these solutions shows exactly how NCERT questions are converted into board exam questions.

Other Class 10 Maths Resources

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FAQs

1. Does the guide include the proofs of theorems for Geometry?

Yes. For chapters like Triangles (Chapter 6) and Circles (Chapter 10), we provide detailed, step-by-step proofs for all major theorems (e.g., Basic Proportionality Theorem). Since these are often asked directly in the 5-mark section of the board paper, we explain the "Given," "To Prove," and "Construction" steps clearly.

2. How are the case study questions for Maths handled?

Following the latest CBSE pattern, we include Case-Based integrated questions for chapters like Applications of Trigonometry and Surface Areas and Volumes. These solutions help you practice extracting mathematical data from real-life scenarios, which now accounts for a significant portion of your marks.

3. Are there tips for solving the Identity problems in Trigonometry?

Trigonometric identities in Chapter 8 can be tricky. Our solutions include "Strategic Hints" for every proof, showing you whether to convert terms into Sine/Cosine or use a specific algebraic identity first to reach the RHS faster.

4. Do the solutions provide calculational shortcuts for Statistics?

Yes. For Chapter 13 (Statistics), we show you how to use the Assumed Mean and Step Deviation methods to simplify large calculations. This reduces the risk of manual errors and helps you finish the paper on time.

5. Do these solutions follow the "Step-Marking" criteria?

Yes. Every solution is written exactly how you should present it in your Board Answer Sheet. We include the formula used, the substitution step, and the final answer with the correct units (cm², m³, etc.), as skipping these can lead to half-mark deductions.

6. How are the Graphs for Linear Equations explained?

For Chapter 3, our solutions don't just give the answer; they provide the coordinate tables and scale suggestions for the graph. We also explain how to interpret the intersection points to find the unique solution, infinitely many solutions, or no solution.

7. Is there a focus on the "Value of π" and "Square Roots"?

In chapters like Area Related to Circles, we specify when to use π = 22/7 versus 3.14 to make calculations easier. We also provide simplified square root values and prime factorization steps for Chapter 1 (Real Numbers) to help with HCF and LCM problems.