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Karnataka SSLC Model Question Paper 2023-24 Mathematics Question Paper with Step-by-Step Solutions

AI-written · numbers calculator-checked on 23 Oct 2026

Step-by-step working for 44 questions of the board's own model paper “S.S.L.C. MODEL QUESTION PAPER – 2023-24, MATHEMATICS, English Medium, subject code 81-E (CCE RF)” (80 marks, 3 hours 15 minutes (including 15 minutes to read the paper)), written by Shishya. Each question is in our own shorter words; open the board's paper for the exact wording, figures and instructions.

The board's paper (PDF) ↗Where the board lists it (kseab.karnataka.gov.in) ↗

How to answer this paper: 38 questions in six parts, all to be answered; Q17, Q21, Q25, Q27, Q28, Q33 and Q36 each have an OR choice (answer one of the two).

How these were checked: the solutions were drafted with AI from the board's English-medium paper (read on 23 Oct 2026). On 23 Oct 2026, every final number was recomputed from the question's own numbers by a calculator check that runs before each release. They were not compared with a board answer key: none was listed beside this paper on the board's page when Shishya read it. Graphs, constructions and diagrams are described step by step, not drawn. If something looks wrong, your teacher and the board's own answer key, where one exists, are the authority.

Free to read: the first 5 solutions are open to everyone; the other 39 open with a free Shishya account (for students 13 and above).

I

  1. Q1

    1 markReal numbers

    Every positive odd integer is of the form (q a positive integer) …

    1. Dividing by 2 leaves remainder 0 (even) or 1 (odd), so an odd integer is 2q + 1.

    Answer: 2q + 1 — option (A).

  2. Q2

    1 markPair of linear equations

    The lines x + 2y = 8 and 2x + 4y = 10 are …

    1. a₁/a₂ = 1/2, b₁/b₂ = 2/4 = 1/2, c₁/c₂ = 8/10 = 4/5.
    2. a₁/a₂ = b₁/b₂ ≠ c₁/c₂: the lines never meet.

    Answer: Parallel to each other — option (D).

  3. Q3

    1 markArithmetic progressions

    The nth term of an A.P. with first term a and common difference d.

    1. Each step adds d; the nth term has had (n − 1) steps from a.

    Answer: aₙ = a + (n − 1)d — option (C).

  4. Q4

    1 markPolynomials

    The sum of the zeroes of p(x) = x² − 2x − 8.

    1. Sum of zeroes = −b/a = −(−2)/1 = 2 (the zeroes are 4 and −2).

    Answer: 2 — option (B).

  5. Q5

    1 markTrigonometry

    If tan θ = 1, find sec θ.

    1. tan θ = 1 means θ = 45°.
    2. sec 45° = 1/cos 45° = √2.

    Answer: √2 — option (C).

  6. Q6

    1 markTriangles (Pythagoras theorem)

    The figure shows ΔPQR with the right angle at R. Which relation is correct?

    Uses a figure printed in the board's paper — open the paper above to see it.

    1. With the right angle at R, the side opposite it, PQ, is the hypotenuse.
    2. Pythagoras: hypotenuse² = sum of the squares of the other two sides.

    Answer: PQ² = PR² + QR² — option (A).

    Steps and answer: open with a free account

  7. Q7

    1 markSurface areas and volumes

    The volume of a cone of radius r and height h.

    1. A cone holds one third of the cylinder with the same base and height.

    Answer: (1/3)πr²h — option (D).

    Steps and answer: open with a free account

  8. Q8

    1 markTrigonometry

    In ΔABC (right angle at B), AB = √3, BC = 1, AC = 2 and ∠ACB = θ. Find θ.

    Uses a figure printed in the board's paper — open the paper above to see it.

    1. At C, the adjacent side is BC = 1 and the hypotenuse is AC = 2.
    2. cos θ = 1/2, so θ = 60° (check: sin θ = √3/2 = AB/AC).

    Answer: 60° — option (B).

    Steps and answer: open with a free account

II

  1. Q9

    1 markReal numbers

    The HCF and LCM of two numbers are 4 and 60. One number is 12. Find the other.

    1. HCF × LCM = product of the numbers: 4 × 60 = 12 × other.
    2. Other = 240 / 12 = 20.

    Answer: 20

    Steps and answer: open with a free account

  2. Q10

    1 markPolynomials

    The degree of g(p) = 7p⁴ − 2p³ + 3p² + p − 3.

    1. The highest power of p is 4.

    Answer: 4

    Steps and answer: open with a free account

  3. Q11

    1 markArithmetic progressions

    The 5th term of the A.P. 3, 1, −1, ….

    1. a = 3, d = 1 − 3 = −2.
    2. a₅ = a + 4d = 3 + 4(−2) = 3 − 8.

    Answer: −5

    Steps and answer: open with a free account

  4. Q12

    1 markQuadratic equations

    Write 2x = 3x² − 5 in standard form.

    1. Bring every term to one side: 3x² − 2x − 5 = 0.

    Answer: 3x² − 2x − 5 = 0

    Steps and answer: open with a free account

  5. Q13

    1 markTrigonometry

    If sin A = 1/2 and cos A = √3/2, find tan A.

    1. tan A = sin A / cos A = (1/2) ÷ (√3/2) = 1/√3.

    Answer: 1/√3

    Steps and answer: open with a free account

  6. Q14

    1 markProbability

    A fair coin is tossed once. Find the probability of a head.

    1. Two equally likely outcomes (H, T); one is a head.

    Answer: 1/2

    Steps and answer: open with a free account

  7. Q15

    1 markCircles

    PA and PB are tangents to a circle with centre O. If ∠AOB = 2∠APB, find ∠APB.

    Uses a figure printed in the board's paper — open the paper above to see it.

    1. ∠OAP = ∠OBP = 90° (tangent ⟂ radius), so in quadrilateral OAPB, ∠AOB + ∠APB = 360° − 180° = 180°.
    2. 2∠APB + ∠APB = 180° → 3∠APB = 180°.

    Answer: ∠APB = 60°

    Steps and answer: open with a free account

  8. Q16

    1 markSurface areas and volumes

    The formula for the curved surface area of the frustum in the figure (radii r₁, r₂, slant height l).

    Uses a figure printed in the board's paper — open the paper above to see it.

    1. The curved surface of a frustum is π times the sum of the radii times the slant height.

    Answer: CSA = π(r₁ + r₂)l

    Steps and answer: open with a free account

III

  1. Q17

    2 marksReal numbers

    Prove that 2 + √3 is irrational.

    1. Suppose 2 + √3 is rational: 2 + √3 = a/b with integers a, b (b ≠ 0).
    2. Then √3 = a/b − 2 = (a − 2b)/b, which is rational, since a − 2b and b are integers.
    3. But √3 is irrational. The contradiction shows the supposition is false.

    Answer: 2 + √3 is irrational — proved

    Steps and answer: open with a free account

  2. Q17 (OR)

    2 marksReal numbers

    Find the HCF of 64 and 332 by Euclid's division algorithm.

    1. 332 = 64 × 5 + 12.
    2. 64 = 12 × 5 + 4.
    3. 12 = 4 × 3 + 0: the remainder is 0, so the last divisor is the HCF.

    Answer: HCF = 4

    Steps and answer: open with a free account

  3. Q18

    2 marksPair of linear equations

    Solve by elimination: 2x + 3y = 14 and 2x + y = 10.

    1. Subtract the second from the first: 2y = 4, so y = 2.
    2. 2x + 2 = 10 → x = 4.

    Answer: x = 4, y = 2

    Steps and answer: open with a free account

  4. Q19

    2 marksArithmetic progressions

    Find the sum of the first 30 terms of 3, 7, 11, … using the formula.

    1. a = 3, d = 4, n = 30.
    2. S₃₀ = 30/2 [2 × 3 + 29 × 4] = 15 × [6 + 116] = 15 × 122.

    Answer: 1830

    Steps and answer: open with a free account

  5. Q20

    2 marksQuadratic equations

    Find the roots of x² − 7x + 12 = 0 using the quadratic formula.

    1. a = 1, b = −7, c = 12; b² − 4ac = 49 − 48 = 1.
    2. x = [7 ± √1] / 2 = (7 + 1)/2 or (7 − 1)/2.

    Answer: x = 4 or x = 3

    Steps and answer: open with a free account

  6. Q21

    2 marksTrigonometry

    Prove sin 30° + cos 60° + tan 45° = sec 60°.

    1. Left side = 1/2 + 1/2 + 1 = 2.
    2. Right side = sec 60° = 1/cos 60° = 2.

    Answer: Both sides = 2 — proved

    Steps and answer: open with a free account

  7. Q21 (OR)

    2 marksTrigonometry

    Prove cos A/(1 + sin A) + (1 + sin A)/cos A = 2 sec A.

    1. Common denominator: [cos²A + (1 + sin A)²] / [(1 + sin A) cos A].
    2. Numerator = cos²A + 1 + 2 sin A + sin²A = 2 + 2 sin A = 2(1 + sin A).
    3. So the sum = 2(1 + sin A) / [(1 + sin A) cos A] = 2/cos A = 2 sec A.

    Answer: = 2 sec A — proved

    Steps and answer: open with a free account

  8. Q22

    2 marksCoordinate geometry

    Find the point dividing the join of (2, 1) and (7, 6) in the ratio 3 : 2.

    1. Section formula: ((m x₂ + n x₁)/(m + n), (m y₂ + n y₁)/(m + n)) with m = 3, n = 2.
    2. x = (21 + 4)/5 = 5, y = (18 + 2)/5 = 4.

    Answer: (5, 4)

    Steps and answer: open with a free account

  9. Q23

    2 marksProbability

    Tokens numbered 1 to 15; one is drawn. Find the probability that it does not bear a prime number.

    1. Primes from 1 to 15: 2, 3, 5, 7, 11, 13 — 6 tokens.
    2. Not prime: 15 − 6 = 9 tokens (1 is not prime).
    3. P = 9/15.

    Answer: 3/5

    Steps and answer: open with a free account

  10. Q24

    2 marksConstructions

    Draw a pair of tangents to a circle of radius 4 cm inclined to each other at 60°.

    1. The angle between the tangents and the angle between the radii to their points of contact add to 180°: so the radii make 180° − 60° = 120°.
    2. Draw a circle of radius 4 cm with centre O; draw radius OA and, at O, a radius OB with ∠AOB = 120°.
    3. At A and B draw lines perpendicular to OA and OB; they meet at P.
    4. PA and PB are the tangents; ∠APB = 60°.

    Answer: Tangents PA, PB with ∠AOB = 120°, so ∠APB = 60°

    Steps and answer: open with a free account

IV

  1. Q25

    3 marksPolynomials

    Divide p(x) = x⁴ − 3x² + 4x + 5 by g(x) = x² − 1; find the quotient and remainder.

    1. x⁴ ÷ x² = x²: subtract x²(x² − 1) = x⁴ − x², leaving −2x² + 4x + 5.
    2. −2x² ÷ x² = −2: subtract −2(x² − 1) = −2x² + 2, leaving 4x + 3.
    3. 4x + 3 has a lower degree than x² − 1, so the division stops.
    4. Check: (x² − 1)(x² − 2) + (4x + 3) = x⁴ − 3x² + 2 + 4x + 3 = x⁴ − 3x² + 4x + 5 ✓.

    Answer: q(x) = x² − 2, r(x) = 4x + 3

    Steps and answer: open with a free account

  2. Q25 (OR)

    3 marksPolynomials

    x³ − 3x² + x + 2 divided by g(x) gives quotient x − 2 and remainder −2x + 4. Find g(x).

    1. Dividend = g(x) × quotient + remainder, so g(x) × (x − 2) = x³ − 3x² + x + 2 − (−2x + 4) = x³ − 3x² + 3x − 2.
    2. Divide x³ − 3x² + 3x − 2 by x − 2: x² first (leaves −x² + 3x − 2), then −x (leaves x − 2), then +1 (leaves 0).

    Answer: g(x) = x² − x + 1

    Steps and answer: open with a free account

  3. Q26

    3 marksQuadratic equations

    A rectangular field's diagonal is 20 m more than its shorter side; the shorter side is 10 m less than the longer side. Find the sides.

    1. Shorter side = x m, longer = (x + 10) m, diagonal = (x + 20) m.
    2. x² + (x + 10)² = (x + 20)² → 2x² + 20x + 100 = x² + 40x + 400 → x² − 20x − 300 = 0.
    3. (x − 30)(x + 10) = 0 → x = 30 (a side cannot be −10).
    4. Check: 30² + 40² = 900 + 1600 = 2500 = 50² ✓.

    Answer: Shorter side 30 m, longer side 40 m

    Steps and answer: open with a free account

  4. Q27

    3 marksCoordinate geometry

    Find the area of ΔPQR with P(1, 6), Q(3, 2), R(10, 8).

    1. Area = ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|.
    2. = ½ |1(2 − 8) + 3(8 − 6) + 10(6 − 2)| = ½ |−6 + 6 + 40| = ½ × 40.

    Answer: 20 square units

    Steps and answer: open with a free account

  5. Q27 (OR)

    3 marksCoordinate geometry

    A(1, 4), B(−2, −2), C(4, −2); AD is the median to BC. Find AD.

    1. D is the midpoint of BC: ((−2 + 4)/2, (−2 − 2)/2) = (1, −2).
    2. AD = √[(1 − 1)² + (4 − (−2))²] = √36.

    Answer: AD = 6 units

    Steps and answer: open with a free account

  6. Q28

    3 marksStatistics

    Find the mean: 0–10: 4, 10–20: 6, 20–30: 17, 30–40: 13, 40–50: 7, 50–60: 3.

    1. Class marks xᵢ: 5, 15, 25, 35, 45, 55; Σfᵢ = 50.
    2. fᵢxᵢ: 20, 90, 425, 455, 315, 165; Σfᵢxᵢ = 1470.
    3. Mean = Σfᵢxᵢ / Σfᵢ = 1470 / 50.

    Answer: Mean = 29.4

    Steps and answer: open with a free account

  7. Q28 (OR)

    3 marksStatistics

    Find the mode: 1–5: 1, 5–10: 2, 10–15: 13, 15–20: 15, 20–25: 7, 25–30: 2.

    1. The highest frequency is 15, so the modal class is 15–20: l = 15, f₁ = 15, f₀ = 13, f₂ = 7, h = 5.
    2. Mode = l + [(f₁ − f₀) / (2f₁ − f₀ − f₂)] × h = 15 + [2 / (30 − 13 − 7)] × 5 = 15 + (2/10) × 5.

    Answer: Mode = 16

    Steps and answer: open with a free account

  8. Q29

    3 marksStatistics

    Draw a 'more than' ogive: 50 or more: 100, 55: 98, 60: 90, 65: 77, 70: 49, 75: 15 farms.

    1. X axis: lower limits 50, 55, 60, 65, 70, 75 (kg/hectare); Y axis: cumulative frequency 0 to 100.
    2. Plot (50, 100), (55, 98), (60, 90), (65, 77), (70, 49), (75, 15).
    3. Join the points with a smooth free-hand curve: it falls from left to right.

    Answer: A falling curve through (50, 100), (55, 98), (60, 90), (65, 77), (70, 49), (75, 15)

    Steps and answer: open with a free account

  9. Q30

    3 marksTriangles (similarity)

    D is on BC with ∠BAC = ∠ADB, BC = 8 cm, AB = 6 cm. Prove area ΔABC / area ΔABD = 16/9.

    Uses a figure printed in the board's paper — open the paper above to see it.

    1. In ΔABC and ΔDBA: ∠BAC = ∠BDA (given) and ∠B is common, so ΔABC ~ ΔDBA (AA).
    2. The ratio of areas of similar triangles is the square of the ratio of corresponding sides: BC ↔ BA.
    3. Area ΔABC / area ΔDBA = (BC/BA)² = (8/6)² = 64/36 = 16/9.

    Answer: Ratio = 16/9 — proved

    Steps and answer: open with a free account

  10. Q31

    3 marksCircles

    Prove that the tangents drawn from an external point to a circle are equal.

    1. Let PQ and PR be tangents from P to a circle with centre O, touching it at Q and R. Join OQ, OR and OP.
    2. ∠OQP = ∠ORP = 90° (a tangent is perpendicular to the radius at the point of contact).
    3. In right triangles OQP and ORP: OQ = OR (radii) and OP is common, so ΔOQP ≅ ΔORP (RHS).
    4. Corresponding parts: PQ = PR.

    Answer: PQ = PR — proved

    Steps and answer: open with a free account

  11. Q32

    3 marksConstructions

    Construct a triangle with sides 5 cm, 6 cm and 9 cm, then a similar triangle with sides 2/3 of these.

    1. Draw BC = 9 cm. With B as centre draw an arc of 5 cm and with C as centre an arc of 6 cm; they meet at A. ΔABC is drawn.
    2. From B draw a ray BX at an acute angle to BC, below BC, and mark 3 equal parts B₁, B₂, B₃ on it.
    3. Join B₃C. Through B₂ draw a line parallel to B₃C, meeting BC at C′.
    4. Through C′ draw a line parallel to CA, meeting BA at A′.
    5. ΔA′BC′ is the required triangle: its sides are 2/3 of 5, 6 and 9 cm (about 3.3 cm, 4 cm and 6 cm).

    Answer: ΔA′BC′ with BC′ = 6 cm, BA′ ≈ 3.3 cm, C′A′ = 4 cm

    Steps and answer: open with a free account

  12. Q33 (OR)

    3 marksAreas related to circles

    ABCD is a rectangle with a semicircle APB on AB. BC = 3 × the radius, and the whole figure APBCDA has area 371 cm². Find the length of the semicircular arc.

    Uses a figure printed in the board's paper — open the paper above to see it.

    1. Let the radius be r: AB = 2r and BC = 3r.
    2. Area = rectangle + semicircle = 2r × 3r + ½πr² = 6r² + (11/7)r² = (53/7)r² (π = 22/7).
    3. (53/7)r² = 371 → r² = 371 × 7/53 = 49 → r = 7 cm.
    4. Arc length = πr = 22/7 × 7.

    Answer: 22 cm

    Steps and answer: open with a free account

V

  1. Q34

    4 marksPair of linear equations

    Solve graphically: x + y = 4 and 2x + y = 7.

    1. x + y = 4: points (0, 4), (4, 0), (2, 2). 2x + y = 7: points (0, 7), (2, 3), (3, 1).
    2. Plot both lines on the same axes; they cross at one point.
    3. Read the crossing point: (3, 1). Check: 3 + 1 = 4 ✓ and 6 + 1 = 7 ✓.

    Answer: x = 3, y = 1

    Steps and answer: open with a free account

  2. Q35

    4 marksArithmetic progressions

    An A.P. of 20 terms: a₁ + a₆ = 0, a₄ = 2, a₅ = 6. Find the A.P. and which term is 62.

    1. d = a₅ − a₄ = 6 − 2 = 4. a₄ = a + 3d → 2 = a + 12 → a = −10.
    2. Check: a₆ = −10 + 20 = 10, and a₁ + a₆ = −10 + 10 = 0 ✓.
    3. The A.P. is −10, −6, −2, 2, 6, ….
    4. aₙ = 62: −10 + (n − 1)4 = 62 → (n − 1)4 = 72 → n = 19.

    Answer: −10, −6, −2, 2, 6, …; 62 is the 19th term

    Steps and answer: open with a free account

  3. Q36

    4 marksSome applications of trigonometry

    From A atop building AD, a car is seen at C and, after moving 500 m towards the building, at B; the two angles of depression are complementary. C to D takes 9 minutes at 100 m/minute. Find the height of the building.

    Uses a figure printed in the board's paper — open the paper above to see it.

    1. CD = 9 × 100 = 900 m, so BD = 900 − 500 = 400 m.
    2. Let the angle of depression to C be θ; to B it is 90° − θ. These equal the angles of elevation at C and B.
    3. tan θ = h/900 and tan(90° − θ) = cot θ = h/400.
    4. Multiply: tan θ × cot θ = 1 = h²/(900 × 400) → h² = 3,60,000 → h = 600.

    Answer: The building is 600 m high

    Steps and answer: open with a free account

  4. Q36 (OR)

    4 marksSome applications of trigonometry

    In ΔABC, AD ⟂ BC, ∠ABC = 60°, ∠ACB = 30°, BC = 36 cm. Find AB, AC and AD.

    Uses a figure printed in the board's paper — open the paper above to see it.

    1. ∠BAC = 180° − 60° − 30° = 90°, so BC is the hypotenuse of ΔABC.
    2. AB = BC cos 60° = 36 × 1/2 = 18 cm; AC = BC sin 60° = 36 × √3/2 = 18√3 cm.
    3. In ΔABD: AD = AB sin 60° = 18 × √3/2 = 9√3 cm.

    Answer: AB = 18 cm, AC = 18√3 ≈ 31.18 cm, AD = 9√3 ≈ 15.59 cm

    Steps and answer: open with a free account

  5. Q37

    4 marksTriangles

    Prove the Basic Proportionality (Thales) theorem.

    1. Statement: a line drawn parallel to one side of a triangle, cutting the other two sides at distinct points, divides them in the same ratio. Given ΔABC, DE ∥ BC with D on AB and E on AC; to prove AD/DB = AE/EC.
    2. Join BE and CD; draw EM ⟂ AB and DN ⟂ AC.
    3. ar(ADE)/ar(BDE) = (½ AD × EM)/(½ DB × EM) = AD/DB, and ar(ADE)/ar(CDE) = (½ AE × DN)/(½ EC × DN) = AE/EC.
    4. ΔBDE and ΔCDE stand on the same base DE between the same parallels DE and BC, so ar(BDE) = ar(CDE).
    5. Therefore AD/DB = AE/EC.

    Answer: AD/DB = AE/EC — proved

    Steps and answer: open with a free account

VI

  1. Q38

    5 marksSurface areas and volumes

    A test tube is a cylinder with a hemispherical bottom; the diameter is 3.5 cm and the total height 17.5 cm. Find its curved surface area and the volume of solution its hemispherical part can hold.

    Uses a figure printed in the board's paper — open the paper above to see it.

    1. r = 3.5/2 = 1.75 cm. Height of the cylinder = 17.5 − 1.75 = 15.75 cm.
    2. CSA = 2πrh + 2πr² = 2πr(h + r) = 2 × 22/7 × 1.75 × (15.75 + 1.75) = 11 × 17.5 = 192.5 cm².
    3. Hemisphere volume = (2/3)πr³ = (2/3) × 22/7 × 1.75³ = (2/3) × 22/7 × 5.359375 ≈ 11.23 cm³.

    Answer: CSA = 192.5 cm²; the hemisphere holds about 11.23 cm³

    Steps and answer: open with a free account

Not solved here

  • Q33 (first choice): The figure's numbers do not fit together: a circle of radius 5 cm cannot have tangents of 8 cm that make an equilateral triangle APB with its chord AB (tangents of 8 cm from a point meet at about 64°, not 60°), so the shaded area depends on which of the given numbers is set aside. The OR choice is solved below.

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