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Class 10 Maths chapter-wise test 2026-27: 14 practice tests with answers

CMDLearn · 4 November 2026

This page has 14 short tests for Class 10 Maths, one for each chapter of the NCERT Mathematics textbook (Reprint 2026-27). Every test is worth 10 marks and takes 25 minutes, and it is built like a small slice of the board paper:

Every question has a worked answer, a marking guide that says where the step marks go, and the textbook pages the idea rests on.

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 copies a textbook question; a few practise a textbook method on new numbers, and none repeats the numbers of our chapter guides, so each test is fresh practice after you finish the chapter. If you find a mistake, please tell us.

How to take a test

1. Finish the chapter and its NCERT exercises first.
2. Set a timer for 25 minutes. Write the answers on paper, with every step, and use π = 22/7 unless the question says otherwise.
3. Mark your work with the marking guide under each answer. Give yourself a step mark only if that step is written down.
4. Score 8 or more: move on. Score 5 to 7: redo the questions you lost marks on in a week. Below 5: go back to the chapter, then try the longer set of 8 practice questions linked under the test.

How the 80 marks are split by unit

CBSE's Mathematics curriculum for Class X, 2026-27, gives the theory paper 80 marks 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 plan your tests by unit: the four Algebra tests together cover a quarter of the paper. The curriculum also sets two limits these tests follow: heights and distances use only 30°, 45° and 60°, and segment areas use only central angles of 60°, 90° and 120°.

The 14 tests

Test 1: Real Numbers (Chapter 1, printed pages 1 to 9)

Q1 (1 mark). The HCF of 2³ × 3² × 7 and 2² × 3⁴ × 5 is: (A) 12 (B) 36 (C) 72 (D) 180

Answer: (B) 36. Take each prime that both numbers share, at the smaller power: 2² × 3² = 36. (7 and 5 are not shared.)

Q2 (2 marks). Find the HCF and LCM of 120 and 168 by prime factorisation, and check that HCF × LCM equals the product of the two numbers.

Answer. 120 = 2³ × 3 × 5 and 168 = 2³ × 3 × 7. HCF = 2³ × 3 = 24. LCM = 2³ × 3 × 5 × 7 = 840. Check: 24 × 840 = 20160, and 120 × 168 = 20160.

Marking: 1 mark for both factorisations, 1 mark for the HCF, LCM and the check.

Q3 (3 marks). Given that √7 is irrational, prove that 3 + 2√7 is irrational.

Answer. Suppose 3 + 2√7 is rational, say 3 + 2√7 = a/b with a and b integers and b ≠ 0. Then √7 = (a − 3b)/(2b). The right side is a ratio of integers, so it is rational, and √7 would be rational. That contradicts the given fact that √7 is irrational, so 3 + 2√7 is irrational.

Marking: 1 mark for the assumption, 1 mark for making √7 the subject, 1 mark for the contradiction and conclusion.

Q4 (4 marks, case study). A school hall is 540 cm wide and 756 cm long. A tiler wants to cover it with identical square tiles, with no tile cut, using the largest tiles possible.

(i) What is the side of the largest tile? (1 mark) (ii) How many such tiles are needed? (1 mark) (iii) The shop only has 36 cm tiles. How many of those are needed, and how many times the number in (ii) is that? (2 marks)

Answer.

• (i) The side must divide both 540 and 756, and be as large as possible: the HCF. 540 = 2² × 3³ × 5 and 756 = 2² × 3³ × 7, so the HCF is 2² × 3³ = 108 cm.
• (ii) 540 ÷ 108 = 5 tiles across and 756 ÷ 108 = 7 tiles along, so 35 tiles.
• (iii) 36 divides both lengths: 540 ÷ 36 = 15 and 756 ÷ 36 = 21, so 315 tiles, which is 315 ÷ 35 = 9 times as many (each side is a third as long, so each 108 cm tile is replaced by 3 × 3 = 9 small ones).

Rests on: pages 2 to 4, section 1.2 (the Fundamental Theorem of Arithmetic, HCF and LCM by prime factorisation), and pages 6 to 9, section 1.3 (proving a number is irrational).

More practice: 8 questions with answers on Real Numbers.

Test 2: Polynomials (Chapter 2, printed pages 10 to 23)

Q1 (1 mark). If 2 is a zero of x² − 7x + k, then k is: (A) 10 (B) −10 (C) 5 (D) 14

Answer: (A) 10. Put x = 2: 4 − 14 + k = 0, so k = 10.

Q2 (2 marks). Find a quadratic polynomial whose zeroes add up to −2 and multiply to −15. Then find the zeroes.

Answer. A quadratic with sum of zeroes S and product P is x² − Sx + P, so the polynomial is x² + 2x − 15. It factorises as (x + 5)(x − 3), so the zeroes are −5 and 3. (Check: −5 + 3 = −2 and −5 × 3 = −15.)

Marking: 1 mark for the polynomial, 1 mark for the zeroes.

Q3 (3 marks). Find the zeroes of 6x² − 7x − 3, then check that their sum and product match what the coefficients predict.

Answer. Split the middle term: 6x² − 9x + 2x − 3 = 3x(2x − 3) + 1(2x − 3) = (3x + 1)(2x − 3). The zeroes are 3/2 and −1/3.

• Sum: 3/2 − 1/3 = 7/6, and −b/a = −(−7)/6 = 7/6.
• Product: (3/2) × (−1/3) = −1/2, and c/a = −3/6 = −1/2.

Marking: 1 mark for the factors, 1 mark for the zeroes, 1 mark for both checks.

Q4 (4 marks, case study). The shape of a garden arch is given by y = −x² + 8x − 12, where x is the distance in metres along the ground from a fixed post and y is the height in metres.

(i) At what values of x does the arch meet the ground? (1 mark) (ii) How wide is the arch at ground level? (1 mark) (iii) The arch is highest halfway between its two feet. Find that height. (2 marks)

Answer.

• (i) The arch meets the ground where y = 0: x² − 8x + 12 = 0, so (x − 2)(x − 6) = 0, giving x = 2 and x = 6.
• (ii) 6 − 2 = 4 m.
• (iii) Halfway is x = (2 + 6) ÷ 2 = 4. Then y = −16 + 32 − 12 = 4 m.

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

More practice: 8 questions with answers on Polynomials.

Test 3: Pair of Linear Equations (Chapter 3, printed pages 24 to 37)

Q1 (1 mark). The pair 3x + ky = 7 and 6x + 10y = 11 has no solution when k is: (A) 10 (B) 2 (C) 5 (D) 20

Answer: (C) 5. No solution needs a₁/a₂ = b₁/b₂ ≠ c₁/c₂. Here 3/6 = 1/2, so k/10 = 1/2 gives k = 5, and 7/11 is not 1/2.

Q2 (2 marks). Solve by substitution: x + 5y = 17 and 4x − y = 5.

Answer. From the first, x = 17 − 5y. Put it in the second: 4(17 − 5y) − y = 5, so 68 − 21y = 5 and y = 3. Then x = 17 − 15 = 2. x = 2, y = 3.

Marking: 1 mark for the substitution, 1 mark for both values.

Q3 (3 marks). At a school canteen, 4 samosas and 3 glasses of juice cost ₹110, and 2 samosas and 5 glasses of juice cost ₹132. Find the price of one samosa and one glass of juice.

Answer. Let a samosa cost ₹s and a juice ₹j. Then 4s + 3j = 110 and 2s + 5j = 132. Double the second: 4s + 10j = 264. Subtract the first: 7j = 154, so j = 22. Then 4s = 110 − 66 = 44, so s = 11. Samosa ₹11, juice ₹22.

Marking: 1 mark for the two equations, 1 mark for eliminating one variable, 1 mark for both prices.

Q4 (4 marks, case study). A water tank already holds some water when a pump is switched on, and the pump adds the same amount every minute. After 6 minutes the tank holds 105 litres; after 10 minutes it holds 165 litres.

(i) Taking a litres at the start and r litres per minute, write the two equations. (1 mark) (ii) Find r. (1 mark) (iii) Find a, and the time at which the tank holds 300 litres. (2 marks)

Answer.

• (i) a + 6r = 105 and a + 10r = 165.
• (ii) Subtract: 4r = 60, so r = 15 litres per minute.
• (iii) a = 105 − 90 = 15 litres. For 300 litres: 15 + 15t = 300, so t = 19 minutes.

Rests on: pages 25 to 27, section 3.2 (consistency and the ratios of coefficients), and pages 30 to 35, section 3.3 (substitution and elimination).

More practice: 8 questions with answers on linear equations.

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

Q1 (1 mark). The equation 7x² − 2x + 1 = 0 has: (A) two distinct real roots (B) two equal real roots (C) no real roots (D) more than two real roots

Answer: (C). The discriminant is b² − 4ac = 4 − 28 = −24, which is negative.

Q2 (2 marks). Find the positive value of k for which 4x² + kx + 9 = 0 has two equal roots, and find that root.

Answer. Equal roots need b² − 4ac = 0: k² − 144 = 0, so k = 12. The equation is 4x² + 12x + 9 = (2x + 3)² = 0, so the root is x = −3/2.

Marking: 1 mark for k, 1 mark for the root.

Q3 (3 marks). For a school function, 143 chairs are set out in equal rows, with 2 fewer rows than there are chairs in each row. Form a quadratic equation and find the number of rows and the chairs in each row.

Answer. Let each row have x chairs, so there are x − 2 rows. Then x(x − 2) = 143, so x² − 2x − 143 = 0. Factorise: (x − 13)(x + 11) = 0, so x = 13 or x = −11. A count cannot be negative, so there are 13 chairs in each row and 11 rows.

Marking: 1 mark for the equation, 1 mark for the factors, 1 mark for rejecting the negative root and stating both numbers.

Q4 (4 marks, case study). A rectangular vegetable plot is 7 m longer than it is wide, and its area is 120 m².

(i) Taking the width as b metres, write the quadratic equation. (1 mark) (ii) Find its discriminant. (1 mark) (iii) Find the width and the length of the plot. (2 marks)

Answer.

• (i) b(b + 7) = 120, so b² + 7b − 120 = 0.
• (ii) D = 7² − 4(1)(−120) = 49 + 480 = 529.
• (iii) √529 = 23, so b = (−7 ± 23) ÷ 2 = 8 or −15. A width cannot be negative, so the width is 8 m and the length is 15 m.

Rests on: page 39, section 4.2 (forming the equation), page 42, section 4.3 (factorisation), and pages 44 to 46, section 4.4 (the discriminant).

More practice: 8 questions with answers on Quadratic Equations.

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

Q1 (1 mark). The 15th term of the AP 7, 11, 15, 19, ... is: (A) 59 (B) 60 (C) 63 (D) 67

Answer: (C) 63. a = 7 and d = 4, so a₁₅ = 7 + 14 × 4 = 63.

Q2 (2 marks). Which term of the AP 5, 12, 19, 26, ... is 166?

Answer. a = 5 and d = 7. Then 5 + (n − 1)7 = 166, so (n − 1)7 = 161, n − 1 = 23 and n = 24. 166 is the 24th term.

Marking: 1 mark for the equation, 1 mark for n.

Q3 (3 marks). Find the sum of all multiples of 7 that lie between 100 and 300.

Answer. The first is 105 and the last is 294, with d = 7. Number of terms: (294 − 105) ÷ 7 + 1 = 28. Sum = (28 ÷ 2)(105 + 294) = 14 × 399 = 5586.

Marking: 1 mark for the first and last terms, 1 mark for n = 28, 1 mark for the sum.

Q4 (4 marks, case study). A new bakery makes 50 cupcakes on its first day and, every day after that, 15 more than the day before.

(i) How many cupcakes does it make on day 10? (1 mark) (ii) How many has it made in total in the first 10 days? (1 mark) (iii) On which day does it first make more than 400 cupcakes? (2 marks)

Answer.

• (i) a₁₀ = 50 + 9 × 15 = 185 cupcakes.
• (ii) S₁₀ = (10 ÷ 2)(50 + 185) = 1175 cupcakes.
• (iii) Need 50 + (n − 1)15 > 400, so (n − 1) > 350 ÷ 15 = 23.3..., so n − 1 = 24 and day 25. Day 24 gives 395 and day 25 gives 410.

Rests on: page 57, section 5.3 (the nth term), and pages 64 to 66, section 5.4 (sum of the first n terms).

More practice: 8 questions with answers on Arithmetic Progressions.

Test 6: Triangles (Chapter 6, printed pages 73 to 98)

Q1 (1 mark). In triangle ABC, D is on AB and E is on AC with DE ∥ BC. If AD = 2.5 cm, DB = 4 cm and AE = 3 cm, then EC is: (A) 3.75 cm (B) 4.8 cm (C) 5 cm (D) 2 cm

Answer: (B) 4.8 cm. By the Basic Proportionality Theorem, AD/DB = AE/EC, so 2.5/4 = 3/EC and EC = 12 ÷ 2.5 = 4.8 cm.

Q2 (2 marks). Triangle ABC is similar to triangle PQR, with AB = 8 cm and PQ = 12 cm. The perimeter of ABC is 30 cm. Find the perimeter of PQR.

Answer. In similar triangles every pair of corresponding sides is in the same ratio, here 12/8 = 3/2. Each side of PQR is 3/2 of the matching side of ABC, so the perimeter is too: 30 × 3/2 = 45 cm.

Marking: 1 mark for the ratio of sides, 1 mark for the perimeter.

Q3 (3 marks). At 4 pm, a boy 1.8 m tall standing on level ground casts a shadow 2.7 m long. At the same moment a tree nearby casts a shadow 42 m long. Find the height of the tree.

Answer. The sun's rays make the same angle with the ground for both, and the boy and the tree both stand at 90° to the ground. So the two right triangles (height, shadow, ray) are similar by the AA criterion. Then height/shadow is equal: h/42 = 1.8/2.7, so h = 42 × 1.8 ÷ 2.7 = 28 m.

Marking: 1 mark for naming the similar triangles and why, 1 mark for the proportion, 1 mark for the height.

Q4 (4 marks, case study). A roof frame is a triangle PQR with a horizontal beam ST, where S is on PQ, T is on PR and ST ∥ QR. PS = 2 m, SQ = 3 m and PT = 1.6 m.

(i) Find TR. (1 mark) (ii) Find PS : PQ. (1 mark) (iii) The base QR is 7.5 m. Find the length of the beam ST. (2 marks)

Answer.

• (i) PS/SQ = PT/TR, so 2/3 = 1.6/TR and TR = 2.4 m.
• (ii) PQ = 2 + 3 = 5 m, so PS : PQ = 2 : 5.
• (iii) ST ∥ QR, so angle PST = angle PQR and angle PTS = angle PRQ (corresponding angles). Triangles PST and PQR are similar (AA), so ST/QR = PS/PQ = 2/5 and ST = 7.5 × 2/5 = 3 m.

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

More practice: 8 questions with answers on Triangles.

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

Q1 (1 mark). The distance between (−2, 7) and (3, −5) is: (A) 12 (B) 13 (C) 17 (D) √119

Answer: (B) 13. √(5² + 12²) = √169 = 13.

Q2 (2 marks). Find the point on the y-axis that is equally far from A(8, 3) and B(−2, 7).

Answer. Let the point be (0, y). Then 8² + (y − 3)² = 2² + (y − 7)², so 64 + y² − 6y + 9 = 4 + y² − 14y + 49. This gives 8y = −20, so y = −5/2. The point is (0, −5/2).

Marking: 1 mark for the equation, 1 mark for the point.

Q3 (3 marks). In what ratio does the y-axis divide the segment joining (−6, 5) and (4, −10)? Find the point where they meet.

Answer. Let the ratio be k : 1. On the y-axis the x-coordinate is 0: (4k − 6)/(k + 1) = 0, so k = 3/2 and the ratio is 3 : 2. The y-coordinate is (3 × (−10) + 2 × 5)/(3 + 2) = −20/5 = −4. The point is (0, −4).

Marking: 1 mark for setting x = 0, 1 mark for the ratio, 1 mark for the point.

Q4 (4 marks, case study). On a park map, the gate is at G(−4, 2) and the fountain is at F(2, 10), with 1 unit = 10 m.

(i) How far is the fountain from the gate, in metres? (1 mark) (ii) A bench is placed halfway along GF. Where is it? (1 mark) (iii) Two lamps divide GF into three equal parts. Find their positions. (2 marks)

Answer.

• (i) GF = √(6² + 8²) = 10 units = 100 m.
• (ii) Mid-point: ((−4 + 2)/2, (2 + 10)/2) = (−1, 6).
• (iii) The lamp nearer the gate divides GF in 1 : 2: ((2 × (−4) + 2)/3, (2 × 2 + 10)/3) = (−2, 14/3). The other divides it in 2 : 1: ((−4 + 2 × 2)/3, (2 + 2 × 10)/3) = (0, 22/3).

Rests on: pages 100 to 102, section 7.2 (the distance formula), and pages 106 to 110, section 7.3 (the section formula and the mid-point).

More practice: 8 questions with answers on Coordinate Geometry.

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

Q1 (1 mark). If A is acute and sin A = 8/17, then cos A is: (A) 15/17 (B) 8/15 (C) 17/15 (D) 15/8

Answer: (A) 15/17. cos² A = 1 − 64/289 = 225/289, so cos A = 15/17.

Q2 (2 marks). Find the value of 3 sin² 60° − 2 cos² 45° + tan² 60°.

Answer. sin 60° = √3/2, cos 45° = 1/√2 and tan 60° = √3. So 3 × 3/4 − 2 × 1/2 + 3 = 9/4 − 1 + 3 = 17/4.

Marking: 1 mark for the three values, 1 mark for the result.

Q3 (3 marks). If 5 tan θ = 12, find (sin θ + cos θ)/(sin θ − cos θ).

Answer. tan θ = 12/5. Divide the top and the bottom by cos θ: (tan θ + 1)/(tan θ − 1) = (17/5)/(7/5) = 17/7. (Or: take the sides 12 and 5, hypotenuse 13, so sin θ = 12/13, cos θ = 5/13, and (17/13)/(7/13) = 17/7.)

Marking: 1 mark for tan θ or the three sides, 1 mark for the substitution, 1 mark for the value.

Q4 (4 marks, case study). A wheelchair ramp rises 0.7 m over a horizontal run of 2.4 m, and the sloping surface is 2.5 m long. Let θ be the angle the ramp makes with the ground.

(i) Find sin θ. (1 mark) (ii) Find tan θ. (1 mark) (iii) Using these sides, show that sec² θ − tan² θ = 1. (2 marks)

Answer.

• (i) sin θ = 0.7/2.5 = 7/25.
• (ii) tan θ = 0.7/2.4 = 7/24.
• (iii) sec θ = 2.5/2.4 = 25/24. Then sec² θ − tan² θ = 625/576 − 49/576 = 576/576 = 1.

Rests on: pages 115 to 118, section 8.2 (the ratios), page 124, section 8.3 (Table 8.1), and page 128, section 8.4 (the identities).

More practice: 8 questions with answers on Trigonometry.

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

Q1 (1 mark). A straight water slide is 14 m long and makes 30° with the ground. How high is its top? (A) 7 m (B) 7√3 m (C) 14√3 m (D) 28 m

Answer: (A) 7 m. Height = 14 × sin 30° = 14 × 1/2 = 7 m.

Q2 (2 marks). A drone hovers straight above a point P on level ground. From a point Q on the ground 40 m from P, the angle of elevation of the drone is 60°. How high is the drone? (Use √3 = 1.732.)

Answer. tan 60° = height/40, so height = 40√3 = 69.28 m.

Marking: 1 mark for the ratio, 1 mark for the height.

Q3 (3 marks). From a balcony 24 m above a straight road, the angles of depression of two buses parked on the same side are 60° and 30°. How far apart are the buses? (Use √3 = 1.732.)

Answer. The angle of depression equals the angle of elevation from the bus. Near bus: distance = 24/tan 60° = 24/√3 = 8√3 m. Far bus: distance = 24/tan 30° = 24√3 m. Gap = 24√3 − 8√3 = 16√3 = 27.71 m.

Marking: 1 mark for each distance, 1 mark for the gap.

Q4 (4 marks, case study). A zip line runs in a straight line from a platform 15 m high to an anchor on level ground, at 30° to the ground.

(i) How long is the zip line? (1 mark) (ii) How far is the anchor from the foot of the platform? (Use √3 = 1.732.) (1 mark) (iii) A rider stops when they are 6 m above the ground. How far along the line have they travelled? (2 marks)

Answer.

• (i) sin 30° = 15/length, so length = 15 ÷ 1/2 = 30 m.
• (ii) tan 30° = 15/distance, so distance = 15√3 = 25.98 m.
• (iii) The rider has come down 15 − 6 = 9 m. Along the line, the drop is (distance travelled) × sin 30°, so distance travelled = 9 ÷ 1/2 = 18 m.

Rests on: pages 133 to 134, section 9.1 (angles of elevation and depression), and page 124, section 8.3 (Table 8.1).

More practice: 8 questions with answers on applications of trigonometry.

Test 10: Circles (Chapter 10, printed pages 144 to 153)

Q1 (1 mark). A point is 29 cm from the centre of a circle of radius 20 cm. A tangent is drawn from it. How long is that tangent, up to where it touches? (A) 9 cm (B) 21 cm (C) 25 cm (D) 49 cm

Answer: (B) 21 cm. The radius meets the tangent at 90°, so the tangent = √(29² − 20²) = √441 = 21 cm.

Q2 (2 marks). Two tangents from an outside point P touch a circle (centre O) at A and B. PA = 12 cm and angle APB = 60°. Find the chord AB and the radius of the circle. (Use √3 = 1.732.)

Answer. PA = PB (tangents from one point are equal), so triangle PAB is isosceles with a 60° angle between the equal sides, which makes it equilateral: AB = 12 cm. OP bisects angle APB, so angle APO = 30°, and in the right triangle OAP the radius OA = 12 × tan 30° = 4√3 = 6.93 cm.

Marking: 1 mark for AB with the reason, 1 mark for the radius.

Q3 (3 marks). All four sides of a quadrilateral ABCD touch a circle. AB = 8 cm, BC = 10 cm and CD = 13 cm. Find AD, giving a reason for each step.

Answer. Let the sides touch the circle at P (on AB), Q (on BC), R (on CD) and S (on DA). Tangents from one point are equal, so AP = AS, BP = BQ, CR = CQ and DR = DS. Adding, AB + CD = (AP + BP) + (CR + DR) = (AS + BQ) + (CQ + DS) = BC + AD. So 8 + 13 = 10 + AD and AD = 11 cm.

Marking: 1 mark for the equal tangents, 1 mark for AB + CD = BC + AD, 1 mark for AD.

Q4 (4 marks, case study). A circular pond has centre O and radius 9 m. A straight footpath touches the pond at T, and a lamp post L stands on the path 12 m from T.

(i) What is the angle OTL, and why? (1 mark) (ii) How far is the lamp post from the centre of the pond? (1 mark) (iii) A second straight path from L touches the pond at S. Find LS and the perimeter of the shape OTLS. (2 marks)

Answer.

• (i) 90°: the tangent at a point is perpendicular to the radius through that point.
• (ii) OL = √(9² + 12²) = √225 = 15 m.
• (iii) Tangents from L are equal, so LS = 12 m. Perimeter = OT + TL + LS + SO = 9 + 12 + 12 + 9 = 42 m.

Rests on: page 146, section 10.2 (Theorem 10.1), and page 149, section 10.3 (Theorem 10.2, equal tangents from an outside point).

More practice: 8 questions with answers on Circles.

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

Q1 (1 mark). The area of a sector of angle 120° in a circle of radius 9 cm, with π = 3.14, is: (A) 28.26 cm² (B) 84.78 cm² (C) 169.56 cm² (D) 254.34 cm²

Answer: (B) 84.78 cm². (120/360) × 3.14 × 81 = 84.78 cm².

Q2 (2 marks). An arc 33 cm long subtends 90° at the centre of a circle. Find the radius and the area of the sector.

Answer. Arc = (90/360) × 2πr, so 33 = (1/4) × 2 × (22/7) × r, which gives r = 21 cm. Sector area = (90/360) × (22/7) × 21 × 21 = 346.5 cm².

Marking: 1 mark for the radius, 1 mark for the area.

Q3 (3 marks). In a circle of radius 18 cm, a chord makes an angle of 60° at the centre. What is the area of the minor segment? (Use π = 3.14 and √3 = 1.73.)

Answer.

• Sector: (60/360) × 3.14 × 18 × 18 = 54 × 3.14 = 169.56 cm².
• The triangle formed by the two radii and the chord has two equal sides and a 60° angle between them, so it is equilateral with side 18 cm: area = (√3/4) × 18² = 81√3 = 140.13 cm².
• Segment = 169.56 − 140.13 = 29.43 cm².

Marking: 1 mark for the sector, 1 mark for the triangle, 1 mark for the segment.

Q4 (4 marks, case study). A lawn sprinkler throws water up to 14 m and turns to and fro through 120°.

(i) What area of lawn does it water? (1 mark) (ii) How long is the curved outer edge of the wet patch? (1 mark) (iii) The gardener sets it to turn through 90° instead. How much less area is watered? (2 marks)

Answer.

• (i) (120/360) × (22/7) × 14 × 14 = 616/3 = 205.33 m².
• (ii) (120/360) × 2 × (22/7) × 14 = 88/3 = 29.33 m.
• (iii) At 90°: (90/360) × (22/7) × 196 = 154 m². Less by 616/3 − 154 = 154/3 = 51.33 m² (the missing 30° sector).

Rests on: pages 154 to 156, section 11.1 (area of a sector, length of an arc, area of a segment).

More practice: 8 questions with answers on areas related to circles.

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

Q1 (1 mark). The curved surface area of a hemisphere of radius 14 cm is: (A) 616 cm² (B) 1232 cm² (C) 1848 cm² (D) 2464 cm²

Answer: (B) 1232 cm². 2πr² = 2 × (22/7) × 196 = 1232 cm².

Q2 (2 marks). A glass paperweight is a cuboid 7 cm × 7 cm × 3 cm with a solid hemisphere of radius 3.5 cm on top. Find its volume.

Answer. Cuboid: 7 × 7 × 3 = 147 cm³. Hemisphere: (2/3) × (22/7) × 3.5³ = 89.83 cm³. Total = 236.83 cm³.

Marking: 1 mark for each part, with the total.

Q3 (3 marks). A wooden cap for a fence post is a cylinder of radius 3 cm and height 8 cm, topped by a cone of the same radius and height 4 cm. Find the area to be painted: the outside, but not the flat bottom. Use π = 22/7.

Answer.

• Slant height of the cone: √(3² + 4²) = 5 cm.
• Cone's curved surface: π × 3 × 5 = 15π. Cylinder's curved surface: 2π × 3 × 8 = 48π.
• Total = 63π = 63 × 22/7 = 198 cm². (The circle where the cone sits on the cylinder is hidden, so it is not counted.)

Marking: 1 mark for the slant height, 1 mark for the two curved surfaces, 1 mark for the total.

Q4 (4 marks, case study). A grain store is a cylinder of radius 3.5 m and height 10 m, with a hemispherical dome of the same radius on top.

(i) Find the volume of the cylindrical part. (1 mark) (ii) Find the total volume of the store. (1 mark) (iii) Find the area of metal sheet needed for the curved wall and the dome. (2 marks)

Answer.

• (i) πr²h = (22/7) × 3.5 × 3.5 × 10 = 385 m³.
• (ii) Dome = (2/3)πr³ = 89.83 m³, so the total is 385 + 89.83 = 474.83 m³.
• (iii) Curved wall 2πrh = 220 m² and dome 2πr² = 77 m², so 297 m².

Rests on: pages 162 to 164, section 12.2 (surface area of two solids joined together), and page 167, section 12.3 (volume of two solids joined together).

More practice: 8 questions with answers on surface areas and volumes.

Test 13: Statistics (Chapter 13, printed pages 171 to 201)

Q1 (1 mark). For a grouped distribution the mean is 24 and the median is 26. Using the empirical relationship, the mode is about: (A) 25 (B) 28 (C) 30 (D) 22

Answer: (C) 30. 3 Median = Mode + 2 Mean, so Mode = 3 × 26 − 2 × 24 = 78 − 48 = 30.

Q2 (2 marks). Find the mode of this distribution.

Answer. The modal class is 28-36 (highest frequency, 15). With l = 28, f₁ = 15, f₀ = 11, f₂ = 9 and h = 8: mode = 28 + (15 − 11)/(2 × 15 − 11 − 9) × 8 = 28 + 4/10 × 8 = 31.2.

Marking: 1 mark for the modal class and values, 1 mark for the mode.

Q3 (3 marks). Twenty students recorded the minutes they spent on homework one evening. Find the mean.

Answer. Class marks: 6, 18, 30, 42. Σfx = 18 + 90 + 240 + 168 = 516 and Σf = 20. Mean = 516 ÷ 20 = 25.8 minutes.

Marking: 1 mark for the class marks, 1 mark for Σfx, 1 mark for the mean.

Q4 (4 marks, case study). A class of 40 students noted how long their journey to school takes.

(i) Which is the median class? (1 mark) (ii) What is the cumulative frequency of the class before it? (1 mark) (iii) Find the median journey time. (2 marks)

Answer.

• (i) n/2 = 20. Cumulative frequencies are 5, 14, 26, ..., so the 20th student is in 10-15.
• (ii) 14.
• (iii) Median = l + ((n/2 − cf)/f) × h = 10 + ((20 − 14)/12) × 5 = 10 + 2.5 = 12.5 minutes.

Rests on: pages 171 to 177, section 13.2 (mean), pages 183 to 185, section 13.3 (mode), pages 188 to 195, section 13.4 (median), and page 197 (the empirical relationship).

More practice: 8 questions with answers on Statistics.

Test 14: Probability (Chapter 14, printed pages 202 to 217)

Q1 (1 mark). A fair die is thrown once. The probability of a number greater than 4 is: (A) 1/6 (B) 1/3 (C) 1/2 (D) 2/3

Answer: (B) 1/3. Two outcomes (5 and 6) out of six.

Q2 (2 marks). A playlist has 6 Hindi, 9 English and 5 Tamil songs, and the shuffle button picks one at random. Find the probability that the song is not English.

Answer. Total = 20 songs. P(English) = 9/20, so P(not English) = 1 − 9/20 = 11/20.

Marking: 1 mark for P(English) or the count of 11, 1 mark for the answer.

Q3 (3 marks). Two spinners each have four equal parts numbered 1, 2, 3 and 4. Both are spun. Find the probability that (a) the sum is 5, (b) both show the same number, (c) the product is even.

Answer. There are 4 × 4 = 16 equally likely outcomes.

• (a) Sum 5: (1, 4), (2, 3), (3, 2), (4, 1), so 4/16 = 1/4.
• (b) Same number: (1, 1), (2, 2), (3, 3), (4, 4), so 4/16 = 1/4.
• (c) The product is odd only when both are odd: 2 × 2 = 4 outcomes. So P(even) = 1 − 4/16 = 3/4.

Marking: 1 mark for each part.

Q4 (4 marks, case study). A school fair sells 250 raffle tickets numbered 1 to 250, and one ticket is drawn for the first prize.

(i) Find the probability that its number is a multiple of 25. (1 mark) (ii) Find the probability that its number is a perfect square. (1 mark) (iii) The first prize ticket was a multiple of 25 and is kept aside. A second ticket is drawn from the rest. Find the probability that it is also a multiple of 25. (2 marks)

Answer.

• (i) 25, 50, ..., 250: 10 tickets. 10/250 = 1/25.
• (ii) 1², 2², ..., 15² = 225 (16² = 256 is too big): 15 tickets. 15/250 = 3/50.
• (iii) 249 tickets are left, and 9 of them are multiples of 25. 9/249 = 3/83.

Rests on: pages 203 to 206, section 14.1 (theoretical probability and the probability of "not E").

More practice: 8 questions with answers on Probability.

After the 14 tests

• Mixed practice: Class 10 Maths important questions 2026-27 has three more questions per chapter, in every board question type.
• Real board papers: Class 10 Maths PYQ chapter wise shows how many marks each chapter got in two recent CBSE papers.
• Timed: the free Class 10 revision papers are laid out in 1, 2, 3 and 5-mark sections, like the board paper.

Where CMDLearn fits

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Sources

All opened and read on 4 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 pages under each test name the sections used).
• CBSE, Mathematics curriculum, Class X, 2026-27 (course structure and unit marks, page 3; limits on heights and distances and on segment areas, in the Unit V and Unit VI rows): cbseacademic.nic.in/web_material/CurriculumMain27/SecPart1/Maths_SecP1X_2026-27.pdf

The unit marks and limits are from the CBSE curriculum, and the methods are from the textbook pages named under each test. The questions, worked answers and marking guides 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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