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AP SSC 2026-27 Mathematics Model Paper 1 Question Paper with Step-by-Step Solutions

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

Step-by-step working for 38 questions of the board's own model paper “SSC PUBLIC EXAMINATIONS 2026-27, MATHEMATICS (MODEL PAPER - 1), ENGLISH VERSION, 15E & 16E” (100 marks, 3 hours 15 minutes (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 (bse.ap.gov.in) ↗

How to answer this paper: 33 questions in four sections, all to be answered. Only Section IV (Q29–33, 8 marks each) has an internal choice: answer (a) or (b) of each.

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.

The PDF opens with the board's blueprint and weightage tables; Model Paper 1 follows them. The board lists a Model Paper 2 on the same page.

Free to read: the first 5 solutions are open to everyone; the other 33 open with a free Shishya account (for students 13 and above).

Section I

  1. Q1

    1 markReal numbers

    A factor tree: y splits into 4 and x, and x splits into 3 and 7. Find x and y.

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

    1. x = 3 × 7 = 21.
    2. y = 4 × x = 4 × 21 = 84.

    Answer: x = 21, y = 84 — option (C).

  2. Q2

    1 markPolynomials

    Which polynomial has zeroes 1 and 2: x² − 3x + 2, x² + 3x + 2, x² − 2x + 3 or x² + 2x − 3?

    1. Zeroes 1 and 2: sum 3, product 2, so the polynomial is x² − 3x + 2.
    2. Check: 1 − 3 + 2 = 0 and 4 − 6 + 2 = 0.

    Answer: x² − 3x + 2 — option (A).

  3. Q3

    1 markPair of linear equations

    x − y = 1 and x + ky = 5 have the solution x = 2, y = 1. Find k.

    1. Put x = 2, y = 1 in x + ky = 5: 2 + k = 5.
    2. k = 3 (and x − y = 2 − 1 = 1 ✓).

    Answer: k = 3 — option (A).

  4. Q4

    1 markQuadratic equations

    Find k if x² + kx + 6 = 0 has equal roots.

    1. Equal roots: discriminant b² − 4ac = 0, so k² − 24 = 0.
    2. k² = 24 → k = ±√24 = ±2√6.

    Answer: k = ±2√6

  5. Q5

    1 markArithmetic progressions

    Which could be an A.P.: 2, 4, 8, 6; 5, 5, 5, 5; 1, 4, 9, 16; 2, 3, 5, 8?

    1. An A.P. has the same difference between consecutive terms.
    2. 5, 5, 5, 5 has difference 0 every time; the others' differences change.

    Answer: 5, 5, 5, 5 — option (B).

  6. Q6

    1 markTriangles (similarity)

    In ΔPQR and ΔTSM, ∠P = 55°, ∠Q = 25°, ∠M = 100°, ∠S = 25°. Which similarity is correct?

    1. ∠R = 180° − 55° − 25° = 100° and ∠T = 180° − 100° − 25° = 55°.
    2. Matching equal angles: P ↔ T (55°), Q ↔ S (25°), R ↔ M (100°).

    Answer: ΔPQR ~ ΔTSM — option (A).

    Steps and answer: open with a free account

  7. Q7

    1 markTrigonometry

    How would you settle a disagreement about whether sin 27° or sin 72° is larger?

    1. As the angle grows from 0° to 90°, sin goes up from 0 to 1, so the larger angle has the larger sine.

    Answer: Explain that sine increases between 0° and 90°, so sin 72° > sin 27° — option (B).

    Steps and answer: open with a free account

  8. Q8

    1 markSome applications of trigonometry

    Order the steps for finding a tower's height: calculate, draw the triangle, identify the measurements, choose the ratio.

    1. First identify what is given (3), then draw and label the right triangle (2), then choose the ratio linking them (4), and finally calculate (1).

    Answer: 3, 2, 4, 1 — option (A).

    Steps and answer: open with a free account

  9. Q9

    1 markCircles

    Draw a rough sketch: a chord of the larger circle is a tangent to a smaller concentric circle.

    1. Draw two circles with the same centre O, one inside the other.
    2. Draw a chord AB of the outer circle that just touches the inner circle at one point P.
    3. Mark OP: it is perpendicular to AB and P is the midpoint of AB.

    Answer: Two concentric circles; chord AB of the outer circle touching the inner circle at P, with OP ⟂ AB

    Steps and answer: open with a free account

  10. Q10

    1 markAreas related to circles

    Area of a circular field of radius 14 m (in m²).

    1. Area = πr² = 22/7 × 14 × 14 = 22 × 28.

    Answer: 616 m² — option (B).

    Steps and answer: open with a free account

  11. Q11

    1 markSurface areas and volumes

    The total surface area of a cylinder of radius r and height h.

    1. TSA = curved surface + two circular ends = 2πrh + 2πr² = 2πr(r + h).

    Answer: 2πr(r + h) — option (B).

    Steps and answer: open with a free account

  12. Q12

    1 markProbability

    A die is thrown once; P(even). Which is true: (A) even and odd are equally likely so P(E) = 0.5, (B) P(E) = n(E)/n(S) = 3/6, both, or neither?

    1. Both statements are correct: 3 of the 6 equally likely faces are even, and 3/6 = 0.5.

    Answer: Both A and B — option (C).

    Steps and answer: open with a free account

Section II

  1. Q13

    2 marksPolynomials

    Form a quadratic polynomial whose zeroes a and b have a + b = 6 and ab = 4.

    1. A quadratic with zeroes a and b is k[x² − (a + b)x + ab].
    2. Take k = 1: x² − 6x + 4.

    Answer: x² − 6x + 4 (or any non-zero multiple of it)

    Steps and answer: open with a free account

  2. Q14

    2 marksQuadratic equations

    Rahul says x² + 5x + 6 = 0 has both roots positive. Do you agree?

    1. x² + 5x + 6 = (x + 2)(x + 3), so the roots are −2 and −3.
    2. Also: the product (6) is positive and the sum (−5) is negative, so both roots must be negative.

    Answer: No — the roots are −2 and −3, both negative

    Steps and answer: open with a free account

  3. Q15

    2 marksTriangles (similarity)

    State the AA criterion for similarity of two triangles.

    1. If two angles of one triangle are equal to two angles of another, the third angles are equal too (angle sum 180°).

    Answer: If two angles of one triangle are respectively equal to two angles of another triangle, the two triangles are similar

    Steps and answer: open with a free account

  4. Q16

    2 marksCoordinate geometry

    A(5, 1), B(1, 5), C(−3, −1) are the vertices of a triangle. Find the length of the median AD.

    1. Group 2: D is the midpoint of BC: ((1 − 3)/2, (5 − 1)/2) = (−1, 2).
    2. Group 3: AD = √[(5 − (−1))² + (1 − 2)²] = √(36 + 1).

    Answer: AD = √37 ≈ 6.08 units

    Steps and answer: open with a free account

  5. Q17

    2 marksTrigonometry

    sin(A + B) = 1 and tan(A − B) = 1/√3, with 0° < A + B ≤ 90° and A > B. Find A and B.

    1. sin(A + B) = 1 → A + B = 90°.
    2. tan(A − B) = 1/√3 → A − B = 30°.
    3. Add: 2A = 120° → A = 60°; B = 30°.

    Answer: A = 60°, B = 30°

    Steps and answer: open with a free account

  6. Q18

    2 marksCircles

    A tangent at P to a circle of radius 5 cm meets a line through the centre O at Q, with OQ = 13 cm. Find PQ.

    1. The tangent is perpendicular to the radius: ∠OPQ = 90°.
    2. PQ = √(OQ² − OP²) = √(169 − 25) = √144.

    Answer: PQ = 12 cm

    Steps and answer: open with a free account

  7. Q19

    2 marksPair of linear equations

    Compare the ratios to decide whether 5x − 3y = 11 and −10x + 6y = 22 are consistent.

    1. a₁/a₂ = 5/(−10) = −1/2, b₁/b₂ = −3/6 = −1/2, c₁/c₂ = 11/22 = 1/2 (both written as ax + by = c).
    2. a₁/a₂ = b₁/b₂ ≠ c₁/c₂: the lines are parallel and have no common solution.

    Answer: Inconsistent (no solution)

    Steps and answer: open with a free account

  8. Q20

    2 marksSome applications of trigonometry

    From the top of a tree of height h on one bank, the two banks of a river are seen at angles of depression θ₁ and θ₂ (θ₁ < θ₂). The river is d wide. Draw a rough diagram.

    1. Draw the tree as a vertical line AB (A the top, B the foot) on one bank; B is on the near bank.
    2. Draw the ground line from B through C (the near bank, close to the tree's foot) to D (the far bank), with CD = d.
    3. Through A draw a horizontal line. The angle of depression to the near bank C is the larger one, θ₂; to the far bank D it is θ₁.
    4. Join AC and AD and label AB = h, CD = d, θ₁ and θ₂.

    Answer: Tree AB = h; near bank C (angle θ₂), far bank D (angle θ₁), CD = d

    Steps and answer: open with a free account

Section III

  1. Q21

    4 marksPolynomials

    The graph of a polynomial cuts the X axis at −1 and 3 (a parabola opening upwards). (i) How many zeroes? (ii) What are they? (iii) −b/a for ax² + bx + c? (iv) The product of the zeroes.

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

    1. (i) The graph meets the X axis at two points, so there are 2 zeroes.
    2. (ii) The zeroes are −1 and 3.
    3. (iii) −b/a = sum of zeroes = −1 + 3 = 2.
    4. (iv) Product = (−1) × 3 = −3.

    Answer: (i) 2 (ii) −1 and 3 (iii) 2 (iv) −3

    Steps and answer: open with a free account

  2. Q22

    4 marksQuadratic equations

    A boat's speed in still water is 11 km/h. It goes 12 km upstream and back in 2 h 45 min. Find the speed of the stream.

    1. Let the stream's speed be s km/h: upstream 11 − s, downstream 11 + s.
    2. 12/(11 − s) + 12/(11 + s) = 2¾ = 11/4.
    3. 12 × 22 / (121 − s²) = 11/4 → 1056 = 11(121 − s²) → 96 = 121 − s² → s² = 25.
    4. s = 5 (a speed cannot be negative).

    Answer: 5 km/h

    Steps and answer: open with a free account

  3. Q23

    4 marksArithmetic progressions

    A spiral of 13 successive semicircles, centres alternately A and B, radii 0.5 cm, 1 cm, 1.5 cm, 2 cm, …. Find its total length (π = 22/7).

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

    1. A semicircle of radius r has length πr; the radii form an A.P. 0.5, 1, 1.5, … with 13 terms.
    2. Sum of radii = 13/2 × (0.5 + 6.5) = 45.5 cm.
    3. Length = π × 45.5 = 22/7 × 45.5.

    Answer: 143 cm

    Steps and answer: open with a free account

  4. Q24

    4 marksTrigonometry

    True or false, with reasons: (A) sin(A + B) = sin A + sin B; (B) sin θ = cos θ for all θ.

    1. (A) Take A = B = 30°: sin 60° = √3/2 ≈ 0.866, but sin 30° + sin 30° = 1. Not equal, so false.
    2. (B) Take θ = 0°: sin 0° = 0 but cos 0° = 1. sin θ = cos θ only at θ = 45° (between 0° and 90°), so false.

    Answer: (A) False (B) False

    Steps and answer: open with a free account

  5. Q25

    4 marksCircles

    Joffer thinks the two tangents from an external point to a circle are not always equal. Resolve his doubt.

    1. Let PQ and PR be the tangents from P to a circle with centre O, touching at Q and R. Join OQ, OR, OP.
    2. ∠OQP = ∠ORP = 90° (tangent ⟂ radius).
    3. In right triangles OQP and ORP: OQ = OR (radii), OP common, so ΔOQP ≅ ΔORP (RHS).
    4. So PQ = PR for every external point: the tangents are always equal.

    Answer: Tangents from an external point are always equal (PQ = PR), so Joffer's idea is not right

    Steps and answer: open with a free account

  6. Q26

    4 marksSurface areas and volumes

    A cube of side 7 cm is topped by a hemisphere. Find the greatest diameter the hemisphere can have and the surface area of the solid.

    1. The hemisphere sits on a face of side 7 cm, so its greatest diameter is 7 cm (r = 3.5 cm).
    2. Surface area = 6 × 7² − πr² (circle covered on top) + 2πr² (hemisphere) = 294 + πr².
    3. πr² = 22/7 × 3.5 × 3.5 = 38.5, so the surface area = 294 + 38.5.

    Answer: Greatest diameter 7 cm; surface area 332.5 cm²

    Steps and answer: open with a free account

  7. Q27

    4 marksStatistics

    Write the formula for the median of grouped data and explain each symbol.

    1. Median = l + [(n/2 − cf) / f] × h.
    2. l = lower limit of the median class (the class whose cumulative frequency first goes past n/2).
    3. n = number of observations (total frequency); cf = cumulative frequency of the class just before the median class.
    4. f = frequency of the median class; h = class size (width).

    Answer: Median = l + [(n/2 − cf)/f] × h, with l, n, cf, f, h as above

    Steps and answer: open with a free account

  8. Q28

    4 marksProbability

    Write four different probability questions about two dice thrown together, like the example given.

    1. Two dice give 6 × 6 = 36 equally likely outcomes; each question below is answered from them (any four correct questions earn the marks).
    2. Q1: P(sum is 8)? (2,6), (3,5), (4,4), (5,3), (6,2): 5/36.
    3. Q2: P(a doublet)? (1,1) … (6,6): 6/36 = 1/6.
    4. Q3: P(sum is 13)? Impossible: 0. Q4: P(product is 6)? (1,6), (2,3), (3,2), (6,1): 4/36 = 1/9.

    Answer: For example: sum 8 → 5/36; doublet → 1/6; sum 13 → 0; product 6 → 1/9

    Steps and answer: open with a free account

Section IV

  1. Q29(a)

    8 marksReal numbers

    Given √5 is irrational, show that 2 + 3√5 is irrational.

    1. Suppose 2 + 3√5 is rational: 2 + 3√5 = a/b with integers a, b (b ≠ 0).
    2. Then 3√5 = a/b − 2 = (a − 2b)/b, so √5 = (a − 2b)/(3b).
    3. The right side is a ratio of integers, i.e. rational — but √5 is irrational (given).
    4. This contradiction shows the supposition is false.

    Answer: 2 + 3√5 is irrational — proved

    Steps and answer: open with a free account

  2. Q29(b)

    8 marksArithmetic progressions

    Find the sum of the first 51 terms of an A.P. whose second and third terms are 14 and 18.

    1. d = 18 − 14 = 4; a = 14 − 4 = 10.
    2. S₅₁ = 51/2 [2 × 10 + 50 × 4] = 51/2 × 220 = 51 × 110.

    Answer: 5610

    Steps and answer: open with a free account

  3. Q30(a)

    8 marksStatistics

    Marks of 40 students: 0–10: 3, 10–20: 5, 20–30: 9, 30–40: 10, 40–50: 8, 50–60: 5. (i) Class marks, (ii) fᵢxᵢ, (iii) the mean by the direct method.

    1. (i) Class marks xᵢ = (lower + upper)/2: 5, 15, 25, 35, 45, 55.
    2. (ii) fᵢxᵢ: 3 × 5 = 15, 5 × 15 = 75, 9 × 25 = 225, 10 × 35 = 350, 8 × 45 = 360, 5 × 55 = 275.
    3. (iii) Σfᵢ = 40, Σfᵢxᵢ = 1300; mean = 1300 / 40.

    Answer: Mean = 32.5 marks

    Steps and answer: open with a free account

  4. Q30(b)

    8 marksTriangles

    State and prove the Basic Proportionality (Thales) theorem.

    1. Statement: if a line is drawn parallel to one side of a triangle to cut the other two sides at distinct points, it divides them in the same ratio. Given ΔABC, DE ∥ BC, D on AB, 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 are on the same base DE between the parallels DE and BC, so ar(BDE) = ar(CDE).
    5. Hence AD/DB = AE/EC.

    Answer: AD/DB = AE/EC — proved

    Steps and answer: open with a free account

  5. Q31(a)

    8 marksCoordinate geometry

    P(2, −3) and Q(10, y) are 10 units apart. Jyothi says y has only one value; Preethi disagrees. Who is right?

    1. PQ² = (10 − 2)² + (y + 3)² = 100 → 64 + (y + 3)² = 100 → (y + 3)² = 36.
    2. y + 3 = 6 or y + 3 = −6, so y = 3 or y = −9.
    3. Both points (10, 3) and (10, −9) are 10 units from P.

    Answer: Preethi is right: y = 3 or y = −9

    Steps and answer: open with a free account

  6. Q31(b)

    8 marksAreas related to circles

    A round table cover of radius 28 cm has six equal designs (the segments between the circle and the six chords of an inscribed regular hexagon). Making them costs ₹0.35 per cm². Find the total cost.

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

    1. The paper gives no values for π and √3; we take π = 22/7 and √3 = 1.7.
    2. Group 1: six equal sectors, each of angle 360° ÷ 6 = 60°.
    3. Group 2: one design = sector − equilateral triangle = (60/360) × 22/7 × 28² − (√3/4) × 28² = 410.67 − 333.2 = 77.47 cm².
    4. Group 3: six designs = 6 × 77.47 = 464.8 cm². Cost = 464.8 × 0.35.

    Answer: ₹162.68 (with π = 22/7, √3 = 1.7)

    Steps and answer: open with a free account

  7. Q32(a)

    8 marksProbability

    One card from a deck of 52. Find P of (i) a numbered card, (ii) a letter card, (iii) not a face card, (iv) a diamond but not a face card.

    1. (i) Numbered cards 2–10 in each of 4 suits: 9 × 4 = 36; P = 36/52 = 9/13.
    2. (ii) Letter cards A, J, Q, K: 4 × 4 = 16; P = 16/52 = 4/13.
    3. (iii) Face cards J, Q, K: 12; not a face card: 40; P = 40/52 = 10/13.
    4. (iv) Diamonds that are not J, Q, K: 13 − 3 = 10; P = 10/52 = 5/26.

    Answer: (i) 9/13 (ii) 4/13 (iii) 10/13 (iv) 5/26

    Steps and answer: open with a free account

  8. Q32(b)

    8 marksSurface areas and volumes

    A solid cylinder (height 2.4 cm, radius 0.7 cm) has a cone of the same height and radius hollowed out. Find the total surface area of what remains, to the nearest cm².

    1. Slant height of the cone: l = √(2.4² + 0.7²) = √(5.76 + 0.49) = √6.25 = 2.5 cm.
    2. Remaining surface = cylinder's curved surface + its bottom circle + the cone's curved surface (inside).
    3. = 2πrh + πr² + πrl = πr(2h + r + l) = 22/7 × 0.7 × (4.8 + 0.7 + 2.5) = 2.2 × 8 = 17.6 cm².

    Answer: 17.6 cm², about 18 cm²

    Steps and answer: open with a free account

  9. Q33(a)

    8 marksPair of linear equations

    Draw the graphs of x + 3y = 6 and 2x − 3y = 12 and find the solution from the graph.

    1. x + 3y = 6: points (0, 2), (6, 0), (3, 1). 2x − 3y = 12: points (0, −4), (6, 0), (3, −2).
    2. Plot both lines on the same axes; they cross at (6, 0).
    3. Check: 6 + 0 = 6 ✓ and 12 − 0 = 12 ✓.

    Answer: x = 6, y = 0

    Steps and answer: open with a free account

  10. Q33(b)

    8 marksSome applications of trigonometry

    From the top of a multi-storeyed building, the top and bottom of an 8 m building are seen at depressions of 30° and 45°. Find the height of the multi-storeyed building and the distance between them.

    1. Let the tall building be H m and the distance x m. Bottom at 45°: tan 45° = H/x → x = H.
    2. Top at 30°: tan 30° = (H − 8)/x → H − 8 = H/√3.
    3. H(1 − 1/√3) = 8 → H = 8√3/(√3 − 1) = 4√3(√3 + 1) = 12 + 4√3.

    Answer: Height = distance = 4(3 + √3) ≈ 18.93 m

    Steps and answer: open with a free account

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