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CBSE Class 10 Maths Standard (041) sample paper 2026–27, with marking scheme

A full practice paper in the 2026–27 board pattern, 80 marks for 3 hours. 38 questions in Sections A to E: 20 MCQs and assertion–reason, 2-, 3- and 5-mark questions, and 3 case studies. Paper AW-10M-BP-70027; it follows the board pattern but is not an official CBSE paper.

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General instructions

  1. This paper has 38 questions. All questions are compulsory.
  2. It is in five sections: A, B, C, D and E.
  3. Section A has 20 multiple-choice questions (18 MCQs and 2 assertion–reason questions) of 1 mark each.
  4. Section B has 5 very short answer questions of 2 marks each.
  5. Section C has 6 short answer questions of 3 marks each.
  6. Section D has 4 long answer questions of 5 marks each.
  7. Section E has 3 case-based questions of 4 marks each, with sub-parts of 1, 1 and 2 marks.
  8. Some questions have an internal choice (marked OR): answer only one of them.
  9. Draw neat figures wherever needed. Take π = 22/7 wherever needed, unless the question says otherwise.
  10. Use of a calculator is not allowed.

The paper

Section A: 20 questions of 1 mark each

  1. Q1. 1 mark

    The LCM of 36 and 54 is:

    1. 1944
    2. 90
    3. 18
    4. 108
  2. Q2. 1 mark

    If −7 is a zero of the polynomial x² + kx + 21, then k is:

    1. 10
    2. −3
    3. −10
    4. −7
  3. Q3. 1 mark

    For what value of k does 4x + 4y = 8 and 12x + 12y = k have infinitely many solutions?

    1. k = −24
    2. k = 8
    3. k = 24
    4. k = 11
  4. Q4. 1 mark

    The roots of x² − x − 12 = 0 are:

    1. real and equal
    2. real and distinct
    3. always zero
    4. not real (no real roots)
  5. Q5. 1 mark

    The sum of the first 26 terms of 5, 3, 1, … is:

    1. −1040
    2. −567
    3. 78
    4. −520
  6. Q6. 1 mark

    If two angles of one triangle equal two angles of another, the triangles are similar by:

    1. RHS
    2. SSS
    3. SAS
    4. AA
  7. Q7. 1 mark

    The distance of (−15, −8) from the origin is:

    1. 7 units
    2. 17 units
    3. 289 units
    4. 23 units
  8. Q8. 1 mark

    The distance between (0, 0) and (3, 4) is:

    1. 7
    2. 5
    3. 25
    4. 1
  9. Q9. 1 mark

    The value of tan 45° tan 30° + sin 60° sin 60° is:

    1. −3/4 + √3/3
    2. 1/√3
    3. 3/4 + √3/3
    4. 3/2 + 2√3/3
  10. Q10. 1 mark

    For an acute angle A, sin A is always:

    1. between 0 and 1
    2. greater than 1
    3. negative
    4. equal to 1
  11. Q11. 1 mark

    A vertical pole 21 m tall casts a shadow 21√3 m long. The angle of elevation of the sun is:

    1. 90°
    2. 30°
    3. 45°
    4. 60°
  12. Q12. 1 mark

    PA and PB are tangents from P to a circle with centre O. If ∠APB = 50°, then ∠AOB is:

    1. 65°
    2. 130°
    3. 50°
    4. 310°
  13. Q13. 1 mark

    In one complete turn, a wheel of radius r moves forward by:

    1. πr
    2. 4πr
    3. πr²
    4. 2πr
  14. Q14. 1 mark

    For the data below, the modal class is: Class: 20–25, 25–30, 30–35, 35–40, 40–45 Frequency: 3, 18, 13, 20, 14

    1. 30–35
    2. 20–25
    3. 25–30
    4. 35–40
  15. Q15. 1 mark

    The mean of 32, 89, 36, 69, 75, 57, 46 is:

    1. 59.71
    2. 50.5
    3. 57.71
    4. 404
  16. Q16. 1 mark

    For the data below, the median class is: Class: 10–20, 20–30, 30–40, 40–50, 50–60, 60–70 Frequency: 6, 4, 19, 10, 13, 12

    1. 20–30
    2. 30–40
    3. 40–50
    4. 10–20
  17. Q17. 1 mark

    A card is drawn from a well-shuffled deck of 52 cards. The probability that it is a face card is:

    1. 3/13
    2. 1/4
    3. 10/13
    4. 3/26
  18. Q18. 1 mark

    A die is thrown once. The probability of getting a prime number is:

    1. 2/3
    2. 1/4
    3. 1/2
    4. 1/3
  19. Q19. 1 mark

    Assertion (A): If HCF(14, 22) = 2, then LCM(14, 22) = 154. Reason (R): For any two positive integers a and b, HCF(a, b) × LCM(a, b) = a + b. Choose the correct option.

    1. Both A and R are true, and R is the correct explanation of A.
    2. Both A and R are true, but R is not the correct explanation of A.
    3. A is true, but R is false.
    4. A is false, but R is true.
  20. Q20. 1 mark

    Assertion (A): 216 is a term of the AP 9, 15, 21, …. Reason (R): A number x is a term of an AP with first term a and common difference d exactly when (x − a)/d is a whole number. Choose the correct option.

    1. Both A and R are true, and R is the correct explanation of A.
    2. Both A and R are true, but R is not the correct explanation of A.
    3. A is true, but R is false.
    4. A is false, but R is true.

Section B: 5 questions of 2 marks each

  1. Q21. 2 marks

    Express 11,979 as a product of its prime factors.

  2. Q22. 2 marks

    Solve: 71x + 69y = 978 and 69x + 71y = 982

  3. Q23. 2 marks

    In △ABC, DE ∥ BC. If AD = x, DB = x − 2, AE = x + 4 and EC = x + 1 (in cm), find x.

    OR

    ABCD circumscribes a circle. AB = 18 cm, BC = 20 cm and CD = 22 cm. Find DA and the perimeter of ABCD.

  4. Q24. 2 marks

    Find the perimeter of a sector of angle 100° in a circle of radius 14 cm (π = 22/7).

    OR

    Find the area of the major sector of a circle of radius 21 cm when the minor sector has an angle of 135° (π = 22/7).

  5. Q25. 2 marks

    A jar has 50 marbles. The probability of drawing a red marble is 2/5. How many red marbles are there?

Section C: 6 questions of 3 marks each

  1. Q26. 3 marks

    The floor of a hall measures 105 cm by 90 cm. It is to be paved with square tiles, all of the same size, with no tile cut. Find the side of the largest tile that can be used, and the number of such tiles needed.

  2. Q27. 3 marks

    If α and β are the zeroes of the quadratic polynomial 2x² − x − 8, find the value of (α − β)².

  3. Q28. 3 marks

    △ABC ~ △PQR. The areas of △ABC and △PQR are 8 cm² and 2 cm² respectively. If BC = 10 cm, find QR.

  4. Q29. 3 marks

    PA and PB are tangents drawn from an external point P to a circle with centre O and radius 18 cm. If PA = 80 cm, find OP and the perimeter of the quadrilateral OAPB.

    OR

    A circle is inscribed in △ABC, touching the sides BC, CA and AB at D, E and F respectively. If AB = 19 cm, BC = 16 cm and CA = 17 cm, find the lengths of AF, BD and CE.

  5. Q30. 3 marks

    A chord of a circular mirror of radius 20 cm subtends an angle of 90° at the centre. Find the area of the corresponding (i) minor segment and (ii) major segment. (Use π = 3.14.)

  6. Q31. 3 marks

    The following table gives the distances (in km) travelled by the members of a cycling club on a Sunday. Find the mode of the data. Class interval (distance in km) and frequency (number of members): 10–15: 16, 15–20: 4, 20–25: 12, 25–30: 10, 30–35: 18, 35–40: 10.

    OR

    The following table gives the marks obtained by the students of a class in a test. Find the mean using the direct method. Class interval (marks) and frequency (number of students): 0–10: 19, 10–20: 15, 20–30: 12, 30–40: 10, 40–50: 19, 50–60: 5.

Section D: 4 questions of 5 marks each

  1. Q32. 5 marks

    Find two consecutive even positive integers, the sum of whose squares is 2452.

    OR

    Taxi charges in a city are made up of a fixed charge together with a charge for each kilometre travelled. For a journey of 7 km the charge is Rs 159, and for a journey of 30 km the charge is Rs 550. Find the fixed charge and the charge per kilometre. How much does a person have to pay for a journey of 18 km?

  2. Q33. 5 marks

    Isha saves Rs 220 in the first month of a year, and in every month after that saves Rs 45 more than in the month before. In how many months will Isha's total savings be Rs 7,175? How much does Isha save in the last of these months?

  3. Q34. 5 marks

    ZTWD is a trapezium with ZT ∥ DW. L and G are points on the non-parallel sides ZD and TW respectively such that LG ∥ ZT. Prove that ZL/LD = TG/GW. Hence, if ZL = 9 cm, LD = 15 cm and TW = 24 cm, find TG and GW.

  4. Q35. 5 marks

    An observer at the top of a lighthouse 66 m above sea level watches a ship sailing straight towards the lighthouse. The angle of depression of the ship changes from 30° to 45° during the observation. Find the distance travelled by the ship during this time, and its distance from the lighthouse at the second observation. (Take √3 = 1.73.)

    OR

    From the top of a building 24 m high, the angle of elevation of the top of a chimney is 45° and the angle of depression of its foot is 30°. The building and the chimney stand on the same level ground. Find the height of the chimney and the distance between the building and the chimney. (Take √3 = 1.73.)

Section E: 3 case-based questions of 4 marks each

  1. Q36. 4 marks

    A town map

    Arjun draws a map of the town on a coordinate grid, with the town hall at the origin and each unit standing for 1 km. The house H is at (−10, −9), the library S is at (−12, −5) and the post office M is at (−15, 1). A straight road runs from H to M, and the main highway of the town runs along the x-axis.

    1. How far is the library from the town hall?

      1 mark

    2. A bus stop is to be built midway between H and M. Find its coordinates.

      1 mark

    3. Using the distance formula, show that H, S and M lie on one straight line.

      2 marks

      OR

      The highway (the x-axis) crosses the road HM. Find the ratio in which it divides HM, and the point where it crosses.

  2. Q37. 4 marks

    A tower seen from a building

    Kabir stands on the flat roof of a building 30 m tall and looks at a mobile tower across the street. The angle of elevation of the top of the tower from the roof is 30°, and the angle of depression of its foot is 60°. The building and the tower stand on the same level ground. Ignore Kabir's own height.

    1. How far is the building from the tower?

      1 mark

    2. How much taller is the tower than the building?

      1 mark

    3. Find the height of the tower. (Take √3 = 1.73.)

      2 marks

      OR

      Find the distance from Kabir to the foot of the tower along the line of sight. (Take √3 = 1.73.)

  3. Q38. 4 marks

    A gas cylinder

    A gas cylinder used in a laboratory is in the shape of a cylinder with a hemisphere at each end. Its total length is 92 cm and its diameter is 42 cm. Kabir works out its surface area and its capacity. (Use π = 22/7.)

    1. Find the length of the cylindrical part.

      1 mark

    2. Find the curved surface area of the cylindrical part.

      1 mark

    3. Find the total surface area of the solid.

      2 marks

      OR

      Find the volume of the solid.

Marking scheme

The answers and the marks for each step or point, as an examiner marks them. Open a section to check a question.

Section A: 20 questions of 1 mark each: answers and marks
  1. Q1

    (d) 108 (LCM = 18 × 2 × 3 = 108)

  2. Q2

    (a) 10 ((−7)² + k × (−7) + 21 = 0, so k = 10)

  3. Q3

    (c) k = 24 (the second equation must be 3 times the first)

  4. Q4

    (b) real and distinct (b² − 4ac = 49 > 0)

  5. Q5

    (d) −520 (26/2 × (2 × 5 + 25 × (−2)) = −520)

  6. Q6

    (d) AA

  7. Q7

    (b) 17 units (√(225 + 64) = 17)

  8. Q8

    (b) 5 (√(9 + 16) = 5)

  9. Q9

    (c) 3/4 + √3/3 (1/√3 + 3/4)

  10. Q10

    (a) between 0 and 1 (the opposite side is shorter than the hypotenuse)

  11. Q11

    (b) 30° (tan θ = 21 ÷ 21√3, so θ = 30°)

  12. Q12

    (b) 130° (∠OAP = ∠OBP = 90°, so ∠AOB = 360° − 180° − 50° = 130°)

  13. Q13

    (d) 2πr (its circumference)

  14. Q14

    (d) 35–40 (the highest frequency is 20)

  15. Q15

    (c) 57.71 (404 ÷ 7 = 57.71)

  16. Q16

    (c) 40–50 (n = 64, n/2 = 32; cumulative frequencies 6, 10, 29, 39, 52, 64)

  17. Q17

    (a) 3/13 (12 such cards out of 52)

  18. Q18

    (c) 1/2 (3 favourable outcomes out of 6)

  19. Q19

    (c) A is true, but R is false. (A is true: LCM = 14 × 22 ÷ 2 = 154. R is false.)

  20. Q20

    (d) A is false, but R is true. (A is false: (216 − 9) ÷ 6 = 69/2, which is not a whole number. R is true.)

Section B: 5 questions of 2 marks each: answers and marks
  1. Q21
    1. divide by primes in turn: 3² × 11³ (2)
  2. Q22
    1. add: x + y = 14; subtract: x − y = −2 (2)
  3. Q23
    1. AD/DB = AE/EC: (x)(x + 1) = (x − 2)(x + 4), which gives x = 8 (2)
    Q23 OR
    1. AB + CD = BC + DA; perimeter = 2(AB + CD) (2)
  4. Q24
    1. 2r + arc = 28 + 24 4/9 (2)
    Q24 OR
    1. major sector angle = 360° − 135° = 225° (2)
  5. Q25
    1. 2/5 × 50 = 20 (2)
Section C: 6 questions of 3 marks each: answers and marks
  1. Q26
    1. Prime factorisation: 105 = 3 × 5 × 7, 90 = 2 × 3² × 5 (1)
    2. Side of the largest tile = HCF = 15 cm (1)
    3. Number of tiles = 7 × 6 = 42 (1)
  2. Q27
    1. α + β = 1/2 and αβ = −4 (1)
    2. Writing (α − β)² = (α + β)² − 4αβ (1)
    3. Value = 65/4 (1)
  3. Q28
    1. ar(ABC)/ar(PQR) = (BC/QR)², for similar triangles (1)
    2. Substituting the given values (1)
    3. QR = 5 cm (1)
  4. Q29
    1. OA ⟂ PA (radius ⟂ tangent), so △OAP is right-angled at A (1)
    2. OP = 82 cm (1)
    3. PB = PA, perimeter = 196 cm (1)
    Q29 OR
    1. Tangents from an external point are equal: AF = AE, BD = BF, CD = CE (1)
    2. Equations formed, AF + BD + CE = 26 (1)
    3. AF = 10 cm, BD = 9 cm, CE = 7 cm (1)
  5. Q30
    1. Area of the sector = (90/360) × 3.14 × 20² = 314 cm² (1)
    2. area of the right triangle = (1/2) × 20 × 20 = 200 cm²; minor segment = 314 − 200 = 114 cm² (1)
    3. Major segment = 3.14 × 20² − 114 = 1,142 cm² (1)
  6. Q31
    1. Modal class = 30–35 (highest frequency 18); l = 30, h = 5, f₁ = 18, f₀ = 10, f₂ = 10 (1)
    2. Mode = l + (f₁ − f₀)/(2f₁ − f₀ − f₂) × h = 30 + (8/16) × 5 (1)
    3. Mode = 32.5 (1)
    Q31 OR
    1. Class marks xᵢ: 5, 15, 25, 35, 45, 55; fᵢxᵢ: 95, 225, 300, 350, 855, 275 (1)
    2. Σfᵢ = 80, Σfᵢxᵢ = 2,100 (1)
    3. Mean = Σfᵢxᵢ/Σfᵢ = 2,100/80 = 26.25 (1)
Section D: 4 questions of 5 marks each: answers and marks
  1. Q32
    1. Forming the equation: Let the integers be x and x + 2: x² + (x + 2)² = 2452, that is, 2x² + 4x − 2448 = 0 (2)
    2. Solving: x = 34 or x = −36 (2)
    3. Rejecting x = −36: the integers are 34 and 36 (1)
    Q32 OR
    1. Let the fixed charge be Rs x and the charge per kilometre be Rs y: x + 7y = 159 and x + 30y = 550 (2)
    2. Subtracting: 23y = 391, so y = 17; then x = 159 − 7 × 17 = 40 (2)
    3. Fixed charge Rs 40, Rs 17 per kilometre; amount for 18 km = Rs 346 (1)
  2. Q33
    1. Forming the equation: Sₙ = n/2 [2a + (n − 1)d] with a = 220, d = 45: n/2 [440 + (n − 1) × 45] = 7175, that is, 9n² + 79n − 2870 = 0 (2)
    2. Solving: n = 14 or n = −205/9 (2)
    3. Rejecting n = −205/9: 14 months; last term Rs 805 (1)
  3. Q34
    1. Construction: join ZW meeting LG at B (1)
    2. BPT in △ZDW (1)
    3. BPT in △WZT and conclusion (1)
    4. TG = 9 cm (1)
    5. GW = 15 cm (1)
  4. Q35
    1. Angles of depression equal the angles of elevation from the ship (1)
    2. First distance = 66√3 m (1)
    3. Second distance = 66 m (1)
    4. Distance travelled = 66(√3 − 1) m (1)
    5. ≈ 48.2 m (√3 = 1.73) (1)
    Q35 OR
    1. Correct figure described: horizontal line from the top of the building (1)
    2. Distance = 24√3 m from the angle of depression (1)
    3. Part above the roof = 24√3 m (1)
    4. Height = 24(√3 + 1) m (1)
    5. Values to one decimal (√3 = 1.73) (1)
Section E: 3 case-based questions of 4 marks each: answers and marks
  1. Q36 (i)
    1. √((−12)² + (−5)²) = √169 — 13 km (1)
    Q36 (ii)
    1. ((−10 + (−15))/2, (−9 + 1)/2) — (−12.5, −4) (1)
    Q36 (iii)
    1. HS = √((−12 − (−10))² + (−5 − (−9))²) = 2√5, SM = 3√5, HM = 5√5 km (1)
    2. HS + SM = 5√5 = HM, so the points are collinear (1)
    Q36 (iii) OR
    1. Let the ratio be k : 1; the y-coordinate (k × 1 + (−9))/(k + 1) = 0 gives k = 9 (1)
    2. Ratio 9 : 1; x = (9 × (−15) + 1 × (−10))/10 = −14.5, so the point is (−14.5, 0) (1)
  2. Q37 (i)
    1. 10√3 m (1)
    Q37 (ii)
    1. 10 m (1)
    Q37 (iii)
    1. Height = 30 + 10 (1)
    2. 40 m (1)
    Q37 (iii) OR
    1. sin 60° = 30/s (1)
    2. 20√3 m ≈ 34.6 m (1)
  3. Q38 (i)
    1. 92 − 2 × 21 — 50 cm (1)
    Q38 (ii)
    1. 2πrh = 2 × 22/7 × 21 × 50 — 6,600 cm² (1)
    Q38 (iii)
    1. TSA = 2πrh + 2 × 2πr² = 6,600 + 5,544 (1)
    2. = 12,144 cm² (1)
    Q38 (iii) OR
    1. Volume = πr²h + (4/3)πr³ = 69,300 + 38,808 (1)
    2. = 108,108 cm³ (1)

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