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Class 10 Maths important questions chapter-wise 2026-27, with marks

CMDLearn · 3 November 2026

This page has three questions for each of the 14 chapters of the NCERT Class 10 Mathematics book (Reprint 2026-27). Each one is labelled the way the board paper labels it: 1 mark, 2 marks, 3 marks, 5 marks, or a 4-mark case study. Every question has a worked answer and the textbook page the idea rests on, and each chapter links to a longer set of 8 practice questions.

These questions are our own, drafted with AI help. Every answer was re-computed by a test script, but no practising teacher has reviewed them yet. None of them is a textbook exercise or a sample-paper question with the numbers changed. If you find a mistake, please tell us.

How the 80 marks are split by unit

CBSE's Mathematics curriculum for 2026-27 gives the theory paper 80 marks, split across seven units (course structure, page 3 of the curriculum PDF). The other 20 marks are internal assessment in school.

CBSE gives marks per unit, not per chapter, so any "chapter-wise weightage" you see elsewhere is someone's estimate. Algebra alone is a quarter of the paper, so its four chapters get the most practice time.

How the 2026-27 sample paper asks questions

The Mathematics Standard (041) sample paper for 2026-27 has 38 questions in five sections:

The three case studies in the 2026-27 sample paper come from a pair of linear equations, circles and statistics, so those three chapters get a case-based question below. Calculators are not allowed, and the paper says to take π as 22/7 unless told otherwise. For a section-by-section walk through the paper, read the CBSE Class 10 Maths sample paper 2026-27 guide; for the syllabus limits, the CBSE Class 10 Maths syllabus 2026-27.

How to use this page: cover the answer, write your own working on paper, then compare. In a 3 or 5-mark answer the marks are for the steps, so write each one.

The questions, chapter by chapter

1. Real Numbers (Chapter 1, printed pages 1 to 9)

Q1 (1 mark, multiple choice). p and q are two different prime numbers. The value of HCF(p²q, pq³) × LCM(p²q, pq³) is: (A) p³q⁴ (B) p²q³ (C) pq (D) p³q³

Answer. For any two positive integers, HCF × LCM equals the product of the two numbers. So the value is p²q × pq³ = p³q⁴, option (A). (Check: the HCF is pq and the LCM is p²q³, and pq × p²q³ = p³q⁴.)

Rests on: page 4, section 1.2.

Q2 (3 marks, short answer). A sweet shop has 168 kaju barfis and 252 laddoos. It packs them in boxes so that every box holds the same number of sweets and holds only one kind. The shop wants as few boxes as possible. Using prime factorisation, find how many sweets go in each box and how many boxes are needed.

Answer. The number in each box must divide both 168 and 252, and for the fewest boxes it must be the largest such number, the HCF.

• 168 = 2³ × 3 × 7 and 252 = 2² × 3² × 7.
• HCF: take each prime the two numbers share, at the lower of its two powers, and multiply: 2² × 3 × 7 = 84.
• Boxes: 168 ÷ 84 = 2 boxes of barfi and 252 ÷ 84 = 3 boxes of laddoos.

So 84 sweets per box, 5 boxes.

Rests on: page 4, section 1.2 (HCF by the prime factorisation method).

Q3 (5 marks, long answer). A number N has the prime factorisation N = 2³ × 3² × 5², so N = 1800.

(a) Find the smallest natural number k for which kN is a perfect square, and find the square root of kN.

(b) Write √N in the form a√b, where a and b are natural numbers and b is as small as possible.

(c) Say whether each of these is rational or irrational: (√3 + 1)² − 2√3; √12 ÷ √3; √2 × √8; √2 + √8.

(d) Can the product of two irrational numbers be rational? Use (c) to answer.

Answer.

• (a) A perfect square has every prime to an even power. In N, 3 and 5 already have even powers, but 2 has the power 3. One more 2 makes it even, so k = 2. Then kN = 2⁴ × 3² × 5² = 3600, and its square root is 2² × 3 × 5 = 60.
• (b) N = (2² × 3² × 5²) × 2 = 900 × 2, so √N = 30√2. b = 2 is the smallest possible, because 2 is the only prime left with an odd power.
• (c) (√3 + 1)² − 2√3 = 3 + 2√3 + 1 − 2√3 = 4, rational. √12 ÷ √3 = √4 = 2, rational. √2 × √8 = √16 = 4, rational. √2 + √8 = √2 + 2√2 = 3√2, which is irrational (the textbook proves that 3√2 is irrational in Example 7).
• (d) Yes. √2 and √8 are both irrational, but √2 × √8 = 4.

Rests on: pages 2 to 3, section 1.2 (the Fundamental Theorem of Arithmetic), and pages 6 to 9, section 1.3 (irrational numbers, Example 7).

More practice: 8 questions with answers on Real Numbers.

2. Polynomials (Chapter 2, printed pages 10 to 23)

Q1 (1 mark, assertion and reason). Assertion (A): The polynomial x² + 4x + 5 has no real zeroes. Reason (R): A polynomial of degree 2 cannot have more than two zeroes. (A) Both A and R are true, and R is the correct explanation of A. (B) Both A and R are true, but R is not the correct explanation of A. (C) A is true, but R is false. (D) A is false, but R is true.

Answer. x² + 4x + 5 = (x + 2)² + 1, which is at least 1 for every real x, so it is never 0: A is true. R is also true. But "at most two" allows two, one or none, so R does not explain why this polynomial has none. Option (B).

Rests on: pages 15 and 17, section 2.2.

Q2 (3 marks, short answer). The zeroes of x² − 9x + k are in the ratio 1 : 2. Find both zeroes and the value of k.

Answer. Let the zeroes be α and 2α.

• Sum of zeroes = −(coefficient of x) ÷ (coefficient of x²): α + 2α = 9, so α = 3.
• The zeroes are 3 and 6.
• Product of zeroes = constant term ÷ coefficient of x²: k = 3 × 6 = 18.

Check: x² − 9x + 18 = (x − 3)(x − 6).

Rests on: page 20, section 2.3.

Q3 (5 marks, long answer). The graph of y = ax² + bx + c cuts the y-axis at (0, −12) and the x-axis at two points, one of which is (−2, 0). The sum of the zeroes of ax² + bx + c is 1. Find (i) the other zero, (ii) the values of a, b and c, and (iii) the product of the zeroes, checking it against c/a.

Answer. (i) A zero is where the graph cuts the x-axis, so −2 is a zero. If the other zero is k, then −2 + k = 1, so k = 3.

(ii) At x = 0, y = c, so c = −12. Product of zeroes = c/a: (−2)(3) = −12/a, so −6a = −12 and a = 2. Sum of zeroes = −b/a: 1 = −b/2, so b = −2. The polynomial is 2x² − 2x − 12 = 2(x + 2)(x − 3).

(iii) Product = (−2)(3) = −6, and c/a = −12/2 = −6. They agree.

Rests on: pages 11 to 12, section 2.2 (zeroes on the graph), and page 20, section 2.3.

More practice: 8 questions with answers on Polynomials.

3. Pair of Linear Equations in Two Variables (Chapter 3, printed pages 24 to 37)

Q1 (1 mark, multiple choice). The lines 3x − ky = 7 and 6x + 4y = 14 are coincident when k equals: (A) 2 (B) −2 (C) 4 (D) −4

Answer. Coincident lines need a₁/a₂ = b₁/b₂ = c₁/c₂. Here a₁/a₂ = 3/6 = 1/2 and c₁/c₂ = 7/14 = 1/2, so −k/4 must be 1/2, giving k = −2, option (B).

Rests on: page 26, section 3.2 (Table 3.1).

Q2 (2 marks, very short answer). A purse holds only ₹2 coins and ₹5 coins: 23 coins in all, worth ₹76. How many coins of each kind are there?

Answer. Let there be x coins of ₹2 and y coins of ₹5. Then x + y = 23 and 2x + 5y = 76. Doubling the first gives 2x + 2y = 46; subtracting, 3y = 30, so y = 10 and x = 13. 13 coins of ₹2 and 10 coins of ₹5 (check: 26 + 50 = 76).

Rests on: page 34, section 3.3.2 (elimination method).

Q3 (4 marks, case-based). A school collects rainwater in an underground tank fed by two pipes, A and B, each giving a steady flow. On Monday, pipe A ran for 3 hours and pipe B for 2 hours, and 1800 litres came in. On Tuesday, pipe A ran for 2 hours and pipe B for 3 hours, and 1700 litres came in. Let pipe A give a litres an hour and pipe B give b litres an hour.

• (i) Write the pair of linear equations for this situation. (1 mark)
• (ii) How many litres an hour do the two pipes give together? (1 mark)
• (iii) (A) Find how many litres an hour each pipe gives. (2 marks)
• OR (iii) (B) On Wednesday, pipe A ran for 4 hours, and 2800 litres came in altogether. For how many hours did pipe B run? (2 marks)

Answer. (i) 3a + 2b = 1800 and 2a + 3b = 1700.

(ii) Adding the two equations: 5a + 5b = 3500, so a + b = 700 litres an hour.

(iii) (A) Subtracting the second equation from the first: a − b = 100. With a + b = 700, this gives 2a = 800, so pipe A gives 400 litres an hour and pipe B gives 300 litres an hour.

(iii) (B) From a + b = 700 and a − b = 100 (subtracting the two equations), a = 400 and b = 300. If pipe B ran for t hours: 4 × 400 + 300t = 2800, so 300t = 1200 and t = 4 hours.

Rests on: pages 30 to 35, section 3.3 (algebraic methods).

More practice: 8 questions with answers on Pair of Linear Equations.

4. Quadratic Equations (Chapter 4, printed pages 38 to 47)

Q1 (1 mark, multiple choice). Which of these quadratic equations has two equal real roots? (A) x² − 6x + 8 = 0 (B) 4x² − 12x + 9 = 0 (C) x² + x + 1 = 0 (D) 2x² − 5x − 3 = 0

Answer. Equal roots need b² − 4ac = 0. The four values are 36 − 32 = 4, 144 − 144 = 0, 1 − 4 = −3 and 25 + 24 = 49. Only (B) gives 0 (its root is 3/2, repeated).

Rests on: page 44, section 4.4.

Q2 (2 marks, very short answer). Riya thinks of a positive number. She subtracts 3 from it, and multiplies the result by 2 more than the number. The answer is 24. Find her number.

Answer. Let the number be x. Then (x − 3)(x + 2) = 24, so x² − x − 30 = 0, which factorises as (x − 6)(x + 5) = 0. So x = 6 or x = −5. The number is positive, so it is 6 (check: 3 × 8 = 24).

Rests on: page 42, section 4.3 (solution by factorisation).

Q3 (5 marks, long answer). At a meeting, every person shook hands once with every other person. (a) There were 66 handshakes in all. Form a quadratic equation and find how many people were at the meeting. (b) Using the discriminant, show that a meeting like this can never have exactly 50 handshakes.

Answer. (a) With n people, each shakes hands with n − 1 others, and each handshake is counted twice that way, so there are n(n − 1)/2 handshakes. Then n(n − 1)/2 = 66, so n² − n − 132 = 0, which factorises as (n − 12)(n + 11) = 0. n cannot be negative, so n = 12 people.

(b) For 50 handshakes, n² − n − 100 = 0. Its discriminant is (−1)² − 4 × 1 × (−100) = 401. 401 is not a perfect square (20² = 400 and 21² = 441), so the roots (1 ± √401)/2 are irrational, and no whole number of people gives 50 handshakes. 50 is impossible.

Rests on: page 39, section 4.2 (forming the equation); page 42, section 4.3; page 44, section 4.4.

More practice: 8 questions with answers on Quadratic Equations.

5. Arithmetic Progressions (Chapter 5, printed pages 49 to 72)

Q1 (1 mark, multiple choice). The nth term of an AP is aₙ = 7 − 3n. Its common difference is: (A) 7 (B) 3 (C) −3 (D) 4

Answer. a₁ = 4 and a₂ = 1, so d = a₂ − a₁ = −3, option (C). In general, aₙ₊₁ − aₙ = −3 for every n.

Rests on: pages 51 and 57, sections 5.2 and 5.3.

Q2 (3 marks, short answer). In one section of a stadium, the front row has 18 seats and each row behind it has 3 seats more than the row in front. The last row has 75 seats. Find the number of rows and the total number of seats in the section.

Answer. The rows form an AP with a = 18 and d = 3.

• Last row: 18 + (n − 1) × 3 = 75, so n − 1 = 19 and n = 20 rows.
• Total: S = n/2 × (a + l) = 20/2 × (18 + 75) = 10 × 93 = 930 seats.

Rests on: page 57, section 5.3, and page 64, section 5.4.

Q3 (5 marks, long answer). A tree nursery plants 50 saplings on day 1 and, each day after that, d more saplings than the day before. In the first 12 days it plants 1260 saplings in all. (a) Find d. (b) On which day does the nursery first plant more than 200 saplings? (c) How many saplings does it plant in the first 20 days?

Answer. (a) S₁₂ = 12/2 × [2 × 50 + 11d] = 6(100 + 11d) = 1260, so 100 + 11d = 210 and d = 10.

(b) aₙ = 50 + (n − 1) × 10 = 10n + 40. On day 16 this is exactly 200, which is not more than 200, so the first day above 200 is day 17 (210 saplings).

(c) S₂₀ = 20/2 × [100 + 19 × 10] = 10 × 290 = 2900 saplings.

Rests on: page 57, section 5.3, and page 64, section 5.4.

More practice: 8 questions with answers on Arithmetic Progressions.

6. Triangles (Chapter 6, printed pages 73 to 98)

Q1 (1 mark, multiple choice). In triangle PQR, a line parallel to QR cuts PQ at M and PR at N. If PM : MQ = 2 : 3, then PN : PR is: (A) 2 : 3 (B) 2 : 5 (C) 3 : 5 (D) 3 : 2

Answer. By the basic proportionality theorem, PN : NR = PM : MQ = 2 : 3. So PN is 2 parts out of 2 + 3 = 5 parts of PR, and PN : PR = 2 : 5, option (B). Option (A) is PN : NR, not PN : PR.

Rests on: page 80, section 6.3 (Theorem 6.1).

Q2 (3 marks, short answer). In triangle ABC, AB = 9 cm, AC = 12 cm and BC = 15 cm. Point D is on AB with AD = 3 cm, and point E is on AC with AE = 4 cm. Show that triangle ADE is similar to triangle ABC, find DE, and say why DE is parallel to BC.

Answer.

• The two triangles share ∠A, and AD/AB = 3/9 = 1/3 while AE/AC = 4/12 = 1/3. So the sides that include the common angle are in the same ratio, and triangle ADE ~ triangle ABC by the SAS criterion.
• In similar triangles all corresponding sides are in the same ratio, so DE/BC = 1/3 and DE = 15 ÷ 3 = 5 cm.
• Similar triangles have equal corresponding angles, so ∠ADE = ∠ABC. These are corresponding angles made by the line AB with DE and BC, so DE ∥ BC.

Rests on: page 90, section 6.4 (Theorem 6.5, SAS), and page 79, section 6.3 (meaning of similar triangles).

Q3 (5 marks, long answer). In triangle ABC, AB = 12 cm and BC = 15 cm. P is the point on AB with AP = 4 cm. A line through P parallel to BC meets AC at Q. A line through Q parallel to AB meets BC at R.

(a) Find PQ. (b) Show that BPQR is a parallelogram, and find BR and RC. (c) Show that CR/CB = CQ/CA.

Answer.

• (a) PQ ∥ BC, so ∠APQ = ∠ABC and ∠AQP = ∠ACB, and triangle APQ ~ triangle ABC (AA). So PQ/BC = AP/AB = 4/12 = 1/3, and PQ = 5 cm.
• (b) PQ ∥ BR (part of BC) and QR ∥ PB (part of AB), so both pairs of opposite sides of BPQR are parallel: BPQR is a parallelogram. Opposite sides of a parallelogram are equal, so BR = PQ = 5 cm and RC = 15 − 5 = 10 cm.
• (c) In triangle APQ ~ triangle ABC, AQ/AC = 1/3, so CQ/CA = 1 − 1/3 = 2/3. Also CR/CB = 10/15 = 2/3. The two ratios are equal, as the basic proportionality theorem predicts for QR ∥ AB in triangle CAB.

Rests on: pages 80 to 81, section 6.3 (Theorem 6.1), and page 87, section 6.4 (AAA and AA criterion).

More practice: 8 questions with answers on Triangles.

7. Coordinate Geometry (Chapter 7, printed pages 99 to 112)

Q1 (1 mark, multiple choice). Which of these points is nearest to the origin? (A) (3, 4) (B) (−4, −4) (C) (0, −6) (D) (5, −2)

Answer. Compare squared distances from (0, 0): 9 + 16 = 25, 16 + 16 = 32, 0 + 36 = 36 and 25 + 4 = 29. The smallest is 25, so the nearest point is (3, 4), option (A), at 5 units.

Rests on: page 101, section 7.2 (distance formula).

Q2 (2 marks, very short answer). Show that the points O(0, 0), A(6, 0) and B(3, 3√3) are the vertices of an equilateral triangle.

Answer. OA = √(6² + 0²) = 6. OB = √(3² + (3√3)²) = √(9 + 27) = √36 = 6. AB = √((6 − 3)² + (0 − 3√3)²) = √(9 + 27) = 6. All three sides are 6 units, so the triangle is equilateral.

Rests on: page 101, section 7.2 (distance formula).

Q3 (5 marks, long answer). On a school map drawn on a grid (1 unit = 10 m), the library is at L(2, 1), the canteen at C(8, 9) and the gate at G(8, 1).

(a) Find the length of the straight path LC, in units and in metres. (b) A water tap T is on LC with LT : TC = 1 : 3. Find T. (c) Use distances to show that triangle LGC has a right angle at G. (d) Is the tap closer to the library or to the gate?

Answer.

• (a) LC = √((8 − 2)² + (9 − 1)²) = √(36 + 64) = √100 = 10 units, that is 100 m.
• (b) By the section formula with m1 : m2 = 1 : 3, T = ((1 × 8 + 3 × 2)/4, (1 × 9 + 3 × 1)/4) = (14/4, 12/4) = (3.5, 3).
• (c) LG = √(6² + 0²) = 6 and GC = √(0² + 8²) = 8. Then LG² + GC² = 36 + 64 = 100 = LC², so by the converse of Pythagoras' theorem ∠LGC = 90°.
• (d) LT = 10 ÷ 4 = 2.5 units. TG = √((8 − 3.5)² + (1 − 3)²) = √(20.25 + 4) = √24.25, which is more than √16 = 4. So the tap is closer to the library (25 m against about 49 m).

Rests on: page 101, section 7.2 (distance formula), and page 107, section 7.3 (section formula).

More practice: 8 questions with answers on Coordinate Geometry.

8. Introduction to Trigonometry (Chapter 8, printed pages 113 to 132)

Q1 (1 mark, assertion and reason). Assertion (A): For every angle A between 0° and 90°, tan A is greater than sin A. Reason (R): tan A = sin A ÷ cos A, and for an angle between 0° and 90°, cos A lies between 0 and 1. (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.

Answer. R is true: cos A is the adjacent side over the hypotenuse, a positive number less than 1 because the hypotenuse is the longest side. Dividing the positive number sin A by a number between 0 and 1 gives a bigger number, so tan A is greater than sin A: A is true, and R is the reason. Option (A). (For example, at 60°: tan 60° = √3 and sin 60° = √3/2.)

Rests on: pages 115 to 118, section 8.2 (the ratios, tan A = sin A ÷ cos A, and the remark that sin A and cos A are less than 1).

Q2 (3 marks, short answer). Using the values for 0°, 30°, 45°, 60° and 90°, find every one of these angles A for which sin A + cos A = 1. Show the working for each angle.

Answer.

• 0°: 0 + 1 = 1. Yes.
• 30°: 1/2 + √3/2 = (1 + √3)/2, which is more than 1 because √3 is more than 1. No.
• 45°: 1/√2 + 1/√2 = 2/√2 = √2, more than 1. No.
• 60°: √3/2 + 1/2, the same sum as for 30°. No.
• 90°: 1 + 0 = 1. Yes.

So A = 0° or A = 90°.

Rests on: page 124, section 8.3 (Table 8.1).

Q3 (5 marks, long answer). (a) Prove that sin⁴ A − cos⁴ A = sin² A − cos² A, and that this also equals 2 sin² A − 1. (b) Use the result to find sin⁴ A − cos⁴ A for A = 30°, 45° and 60°. (c) For which of these three angles is it zero?

Answer.

• (a) sin⁴ A − cos⁴ A is a difference of two squares: (sin² A − cos² A)(sin² A + cos² A). Since sin² A + cos² A = 1, it equals sin² A − cos² A. Replacing cos² A by 1 − sin² A gives sin² A − (1 − sin² A) = 2 sin² A − 1.
• (b) A = 30°: 2 × 1/4 − 1 = −1/2. A = 45°: 2 × 1/2 − 1 = 0. A = 60°: 2 × 3/4 − 1 = 1/2.
• (c) It is zero at A = 45°, where sin A = cos A.

Rests on: page 128, section 8.4 (sin² A + cos² A = 1), and page 124, section 8.3 (Table 8.1).

More practice: 8 questions with answers on Introduction to Trigonometry.

9. Some Applications of Trigonometry (Chapter 9, printed pages 133 to 143)

Q1 (1 mark, multiple choice). From a window 10 m above level ground, the angle of depression of a parked car is 30°. The distance of the car from the foot of the building is: (A) 10 m (B) 10√3 m (C) 10/√3 m (D) 20 m

Answer. The angle of depression from the window equals the angle of elevation of the window from the car, 30°. So tan 30° = 10 ÷ d, and d = 10 ÷ (1/√3) = 10√3 m, option (B). 20 m is the slant distance from window to car, not the ground distance.

Rests on: page 134, section 9.1 (angle of depression).

Q2 (3 marks, short answer). A glass lift runs straight up the outside wall of an office building. Neha stands on level ground 30 m from the foot of the wall. As she watches, the angle of elevation of the lift rises from 30° to 60°. How far did the lift travel? (Take her eye at ground level.)

Answer.

• First height: h1 = 30 × tan 30° = 30/√3 = 10√3 m.
• Second height: h2 = 30 × tan 60° = 30√3 m.
• Distance travelled = h2 − h1 = 30√3 − 10√3 = 20√3 m (about 34.6 m).

Rests on: page 133, section 9.1 (angle of elevation), and page 124, section 8.3 (Table 8.1).

Q3 (5 marks, long answer). A hill road is built in two straight parts. The first part is 400 m long and rises at 30° to the horizontal. The second part is 100√2 m long and rises at 45° to the horizontal.

(a) How much height does a car gain on each part, and in total? (b) How far does it move horizontally in total? (c) A new road would rise at a steady 30° all the way. How long must it be to gain the same total height?

Answer.

• (a) First part: rise = 400 × sin 30° = 400 × 1/2 = 200 m. Second part: rise = 100√2 × sin 45° = 100√2 × 1/√2 = 100 m. Total rise = 300 m.
• (b) First part: horizontal = 400 × cos 30° = 400 × √3/2 = 200√3 m. Second part: horizontal = 100√2 × cos 45° = 100 m. Total = (200√3 + 100) m, about 446 m.
• (c) For a 30° road, rise = length × 1/2, so length = 300 ÷ (1/2) = 600 m.

Rests on: page 133, section 9.1 (heights and distances with a right triangle), and page 124, section 8.3 (Table 8.1).

More practice: 8 questions with answers on Some Applications of Trigonometry.

10. Circles (Chapter 10, printed pages 144 to 153)

Q1 (1 mark, multiple choice). A circle has radius 4 cm. A straight line is drawn so that its distance from the centre is exactly 4 cm. The line is: (A) a secant (B) a tangent (C) a line that misses the circle (D) a line through the centre

Answer. The foot of the perpendicular from the centre to the line is 4 cm away, so it lies on the circle. Every other point of the line is farther than 4 cm, so it lies outside. The line meets the circle at exactly one point: a tangent, option (B).

Rests on: page 146, section 10.2 (Theorem 10.1 and its proof).

Q2 (2 marks, very short answer). A circle sits inside triangle ABC and touches AB at P, BC at Q and CA at R. AB = 10 cm, BC = 11 cm and CA = 9 cm. Find AP, BQ and CR.

Answer. Tangents from one outside point are equal: AP = AR = x, BP = BQ = y, CQ = CR = z. Then x + y = 10, y + z = 11 and z + x = 9. Adding, 2(x + y + z) = 30, so x + y + z = 15. So x = 15 − 11 = 4, y = 15 − 9 = 6, z = 15 − 10 = 5: AP = 4 cm, BQ = 6 cm, CR = 5 cm.

Rests on: page 149, section 10.3 (Theorem 10.2).

Q3 (4 marks, case-based). A park has a circular pond with centre O and radius 12 m. Four straight hedges are planted round it, and each one touches the edge of the pond: AB touches it at P, BC at Q, CD at R and DA at S. The gardener measures AP = 16 m, BQ = 9 m and CR = 12 m, and the hedge CD is 24 m long.

• (i) What is the angle between the radius OP and the hedge AB? (1 mark)
• (ii) How long is the hedge AB? (1 mark)
• (iii) (A) Find the total length of the four hedges. (2 marks)
• OR (iii) (B) How far is the corner A from the centre O? (2 marks)

Answer.

• (i) A tangent is perpendicular to the radius at the point of contact: 90°.
• (ii) The two tangents from B are equal, so BP = BQ = 9 m. AB = AP + PB = 16 + 9 = 25 m.
• (iii) (A) Tangents from each corner are equal. DR = CD − CR = 24 − 12 = 12 m, so DS = 12 m, and AS = AP = 16 m, so DA = 12 + 16 = 28 m. CQ = CR = 12 m, so BC = 9 + 12 = 21 m. Total = 25 + 21 + 24 + 28 = 98 m.
• (iii) (B) OP is perpendicular to AB, so triangle OPA has a right angle at P. OA = √(12² + 16²) = √(144 + 256) = √400 = 20 m.

Rests on: page 146, section 10.2 (Theorem 10.1), and page 149, section 10.3 (Theorem 10.2).

More practice: 8 questions with answers on Circles.

11. Areas Related to Circles (Chapter 11, printed pages 154 to 160)

Q1 (1 mark, multiple choice). The radius of a circle is doubled, and the angle of a sector in it is halved. The area of the new sector, compared with the old one, is: (A) the same (B) twice as large (C) half as large (D) four times as large

Answer. Sector area = (θ/360) × πr². Doubling r multiplies the area by 4; halving θ divides it by 2. So the area is 4 ÷ 2 = 2 times as large: (B).

Rests on: page 155, section 11.1 (area of a sector of angle θ).

Q2 (2 marks, very short answer). A sector of a circle of radius 21 cm has the same area as a sector of angle 40° cut from a circle of radius 42 cm. Find the angle of the first sector.

Answer. Equal areas: (θ/360) × π × 21² = (40/360) × π × 42². Cancel π/360: 441θ = 40 × 1764, so θ = 40 × 4 = 160°. (π never has to be put in.)

Rests on: page 155, section 11.1.

Q3 (5 marks, long answer). Two garden sprinklers stand at the same point O. Sprinkler P throws water 14 m and turns through 90°. Sprinkler Q throws water 7 m and turns through 270°. Each wet patch is a sector with its centre at O. Taking π = 22/7:

(i) Find the area each sprinkler wets, and say which wets more.

(ii) Find the length of the curved edge of each wet region.

(iii) Through what angle would Q have to turn to wet exactly as much ground as P?

(iv) The gardener runs a hose along the whole boundary of each wet region (the curved edge and the two straight edges). Which region needs the longer hose, and by how much?

Answer.

• (i) P: (90/360) × (22/7) × 14 × 14 = 154 m². Q: (270/360) × (22/7) × 7 × 7 = 115.5 m². P wets more, by 38.5 m².
• (ii) P: (90/360) × 2 × (22/7) × 14 = 22 m. Q: (270/360) × 2 × (22/7) × 7 = 33 m.
• (iii) (θ/360) × (22/7) × 7 × 7 = 154 gives (θ/360) × 154 = 154, so θ = 360°: Q would have to turn all the way round. (A full circle of radius 7 m has the same area as a quarter circle of radius 14 m.)
• (iv) P: 22 + 14 + 14 = 50 m. Q: 33 + 7 + 7 = 47 m. P's region needs 3 m more hose, even though Q's curved edge is longer.

Rests on: page 155, section 11.1 (area of a sector and length of an arc of angle θ).

More practice: 8 questions with answers on Areas Related to Circles.

12. Surface Areas and Volumes (Chapter 12, printed pages 161 to 170)

Q1 (1 mark, multiple choice). Two identical solid hemispheres of radius r are glued together along their flat faces. The area of the outside of the solid they make is: (A) 2πr² (B) 3πr² (C) 4πr² (D) 6πr²

Answer. The two flat circles are now inside the solid, so only the two curved surfaces show: 2πr² + 2πr² = 4πr² (the solid is a sphere). (C). Option (D) is the trap of adding both hemispheres' total surface areas (3πr² each).

Rests on: page 162, section 12.2 (add only the surfaces you can see).

Q2 (2 marks, very short answer). A paperweight is a solid cube of edge 7 cm with a solid cone standing on its top face. The base of the cone is the largest circle that fits on that face, and the cone is 6 cm tall. Taking π = 22/7, find the volume of the paperweight.

Answer. The largest circle on a 7 cm face has radius 3.5 cm. Cube: 7³ = 343 cm³. Cone: (1/3) × (22/7) × 3.5 × 3.5 × 6 = 77 cm³. Total = 343 + 77 = 420 cm³.

Rests on: page 167, section 12.3 (volume of a combination = sum of the parts).

Q3 (5 marks, long answer). A wooden block is a cuboid 14 cm long, 10 cm wide and 7 cm high. A cylindrical hole of diameter 7 cm is drilled straight down through it, from the centre of the top face to the bottom face. Taking π = 22/7, find (i) the volume of wood left and (ii) the total surface area of the block after drilling, inside of the hole included.

Answer.

• Hole: radius 3.5 cm, height 7 cm.
• (i) Cuboid volume = 14 × 10 × 7 = 980 cm³. Hole = (22/7) × 3.5 × 3.5 × 7 = 269.5 cm³. Wood left = 980 − 269.5 = 710.5 cm³.
• (ii) Cuboid surface = 2(14 × 10 + 10 × 7 + 14 × 7) = 2(140 + 70 + 98) = 616 cm². The hole removes a circle from the top and one from the bottom: 2 × (22/7) × 3.5 × 3.5 = 77 cm². It adds the inside wall, a curved surface: 2 × (22/7) × 3.5 × 7 = 154 cm². Total = 616 − 77 + 154 = 693 cm².

Rests on: pages 162 and 167, sections 12.2 and 12.3.

More practice: 8 questions with answers on Surface Areas and Volumes.

13. Statistics (Chapter 13, printed pages 171 to 201)

Q1 (1 mark, multiple choice). Mean by the step-deviation method uses u = (x − a)/h, where x is the class mark. With the classes 10-20, 20-30, ..., 50-60, the assumed mean a = 35 and h = 10, the value of u for the class 50-60 is: (A) −2 (B) 1.5 (C) 2 (D) 20

Answer. Class mark of 50-60 = 55. u = (55 − 35)/10 = 2: (C). (1.5 comes from using the lower limit 50 instead of the class mark.)

Rests on: page 177, section 13.2 (step-deviation method).

Q2 (3 marks, short answer). For the table below, Ravi says: "The modal class is 30-40, so the mode is its class mark, 35." Find the mode correctly and say why it is not 35.

Answer. Modal class 30-40: l = 30, h = 10, f₁ = 15, f₀ = 9, f₂ = 11. Mode = l + (f₁ − f₀)/(2f₁ − f₀ − f₂) × h = 30 + 6/(30 − 9 − 11) × 10 = 30 + 6 = 36. It is not the class mark because the formula weighs the neighbours: the class after (11) is fuller than the class before (9), so the mode is pulled above the middle of the class.

Rests on: page 184, section 13.3 (the mode formula).

Q3 (4 marks, case-based). A cycle-repair shop noted how long each of its last 40 repairs took.

• (i) How many repairs took an hour or more? (1 mark)
• (ii) Which is the median class? (1 mark)
• (iii) (A) Find the median time. (2 marks)
• OR (iii) (B) Find the mean time, using class marks. (2 marks)

Answer.

• (i) 10 + 7 = 17.
• (ii) Cumulative frequencies: 4, 11, 23, 33, 40. n/2 = 20, and the first cumulative frequency above 20 is 23, so the median class is 40-60.
• (iii)(A) l = 40, cf = 11, f = 12, h = 20. Median = 40 + (20 − 11)/12 × 20 = 40 + 15 = 55 minutes.
• (iii)(B) Class marks 10, 30, 50, 70, 90. Σfx = 40 + 210 + 600 + 700 + 630 = 2180. Mean = 2180 ÷ 40 = 54.5 minutes.

Rests on: pages 173 and 193, sections 13.2 and 13.4 (class marks, the mean, cumulative frequency and the median formula).

More practice: 8 questions with answers on Statistics.

14. Probability (Chapter 14, printed pages 202 to 217)

Q1 (1 mark, assertion-reason). A number is picked at random from 1 to 25. Assertion (A): The probability that it is a prime number is 2/5. Reason (R): A prime number has exactly two factors, so 1 is not a prime number. (A) Both A and R are true, and R is the correct explanation of A. (B) Both A and R are true, but R is not the correct explanation of A. (C) A is true, but R is false. (D) A is false, but R is true.

Answer. The primes from 1 to 25 are 2, 3, 5, 7, 11, 13, 17, 19, 23: nine of them. P(prime) = 9/25, not 2/5 (10/25 would need 1 to count). A is false and R is true: (D).

Rests on: page 203, section 14.1 (P(E) = favourable outcomes ÷ all outcomes, for equally likely outcomes).

Q2 (2 marks, very short answer). Three school buses, red, blue and green, stop at Asha's corner. She takes whichever comes first, and each is equally likely to come first on any day. Find the probability that she takes a bus of the same colour on Monday and Tuesday.

Answer. Monday-Tuesday pairs: 3 × 3 = 9 equally likely outcomes (RR, RB, RG, BR, BB, BG, GR, GB, GG). Same colour: RR, BB, GG, which is 3. P = 3/9 = 1/3.

Rests on: page 203, section 14.1 (equally likely outcomes; listing pairs as the book does for two dice on page 212).

Q3 (5 marks, long answer). A jar holds red, green and yellow sweets only. One sweet is taken at random. P(red) = 1/5 and P(green) = 1/2. There are 12 yellow sweets. (i) Find P(yellow). (ii) How many sweets are in the jar? (iii) How many are red and how many green? (iv) 10 more red sweets are added. What is P(red) now? (v) Starting again from the original jar, how many green sweets must be taken out so that P(green) becomes 1/3?

Answer.

• (i) P(yellow) = 1 − 1/5 − 1/2 = 3/10.
• (ii) 3/10 of the jar is 12 sweets, so the jar holds 12 × 10/3 = 40.
• (iii) Red = 40/5 = 8, green = 40/2 = 20 (check: 8 + 20 + 12 = 40).
• (iv) 18 red out of 50: P(red) = 18/50 = 9/25.
• (v) Take out g green: (20 − g)/(40 − g) = 1/3, so 60 − 3g = 40 − g, g = 10 (then 10 green out of 30).

Rests on: pages 203 and 205, section 14.1 (the probabilities of all the elementary events add up to 1; P(E) from counts).

More practice: 8 questions with answers on Probability.

Practise more

• Every chapter above links to its own set of 8 practice questions with answers.
• Timed: the free Class 10 revision papers are laid out in 1, 2, 3 and 5-mark sections, like the board paper.
• Planning the year: Class 10 boards, a 60-day revision plan.

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Sources

All opened and read on 3 November 2026. Page numbers are the printed page numbers of the textbook.

• NCERT, Mathematics, Textbook for Class X (Reprint 2026-27), Chapters 1 to 14, at ncert.nic.in/textbook/pdf/jemh101.pdf to ncert.nic.in/textbook/pdf/jemh114.pdf (the page under each answer names the section used).
• CBSE, Mathematics curriculum, Class X, 2026-27 (course structure and unit marks, page 3; internal assessment): cbseacademic.nic.in/web_material/CurriculumMain27/SecPart1/Maths_SecP1X_2026-27.pdf
• CBSE, Class X Mathematics Standard (041) sample question paper for 2026-27 (general instructions, sections A to E, case studies Q36 to Q38): cbseacademic.nic.in/web_material/SQP/ClassX_2026_27/MathsStandard-SQP.pdf

The unit marks and the paper's layout are from the CBSE documents above, and the methods are from the textbook pages named under each answer. The questions and worked answers are our own, and every answer was re-computed by script. This guide is general information; follow your teacher and the official CBSE notices.

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