Tamil Nadu SSLC March 2026 Mathematics Question Paper with Step-by-Step Solutions

1. A has 3 rows and 4 columns, so its order is 3 × 4.
2. The transpose swaps rows and columns: 4 × 3.

Answer: 4 × 3 — option (d).

1. The tops differ in height by 11 − 6 = 5 m and are 12 m apart horizontally.
2. Distance = √(12² + 5²) = √(144 + 25) = √169.

Answer: 13 m — option (a).

1. Add the equations: 4x = 12, so x = 3.
2. y = 8 − 3 = 5.

Answer: (3, 5) — option (c).

1. Every point on x = 11 has x-coordinate 11, whatever y is: a vertical line, 11 units right of the Y axis.

Answer: Parallel to the Y axis — option (b).

1. Square both sides: tan²θ + 2 tan θ cot θ + cot²θ = 4.
2. tan θ × cot θ = 1, so tan²θ + cot²θ + 2 = 4.

Answer: 2 — option (a).

1. Volumes are equal: (4/3)πx³ = (1/3)πx²h.
2. Cancel (1/3)πx²: 4x = h.

Answer: 4x cm — option (c).

1. (4/3)πr₁³ = 8 × (4/3)πr₂³, so r₁³ = 8r₂³.
2. Take cube roots: r₁ = 2r₂.

Answer: 2 : 1 — option (a).

1. σ = √(Σx²/n − (Σx/n)²).
2. Σx²/n = 20 and Σx/n = 4, so σ = √(20 − 16) = √4.

Answer: 2 — option (b).

1. Total marbles = p + q + r; red marbles = p.
2. P(red) = favourable / total.

Answer: p / (p + q + r) — option (b).

1. R = {(0, 3), (1, 4), (2, 5), (3, 6), (4, 7), (5, 8)}.
2. Domain = the first coordinates; range = the second coordinates.

Answer: Domain = {0, 1, 2, 3, 4, 5}; range = {3, 4, 5, 6, 7, 8}

1. A function from X to Y must give every element of X exactly one image in Y.
2. −5, 1 and 3 each have one image, but 4 ∈ X has no image in R.

Answer: No — 4 has no image, so R is not a function from X to Y

1. 71 ≡ x (mod 8) means x is the remainder when 71 is divided by 8.
2. 71 = 8 × 8 + 7.

Answer: x = 7

1. This is a G.P. with a = 9 and r = 3/9 = 1/3; |r| < 1, so the sum to infinity exists.
2. S∞ = a / (1 − r) = 9 / (2/3) = 27/2.

Answer: 27/2 = 13.5

1. x² − 16 = (x + 4)(x − 4) and x² + 8x + 16 = (x + 4)².
2. Cancel one (x + 4) (for x ≠ −4).

Answer: (x − 4) / (x + 4)

1. Let the number be x: x − 1/x = 24/5.
2. Multiply by 5x: 5x² − 5 = 24x → 5x² − 24x − 5 = 0.
3. (5x + 1)(x − 5) = 0 → x = 5 or x = −1/5.

Answer: x = 5 or x = −1/5

1. Row i = 1: |1 − 2| = 1, |1 − 4| = 3, |1 − 6| = 5.
2. Row i = 2: |2 − 2| = 0, |2 − 4| = 2, |2 − 6| = 4.
3. Row i = 3: |3 − 2| = 1, |3 − 4| = 1, |3 − 6| = 3.

Answer: Rows (1 3 5), (0 2 4), (1 1 3)

1. DB = 5.6 − 1.4 = 4.2 cm and EC = 7.2 − 1.8 = 5.4 cm.
2. AD/DB = 1.4/4.2 = 1/3 and AE/EC = 1.8/5.4 = 1/3.
3. AD/DB = AE/EC, so by the converse of the Basic Proportionality Theorem, DE ∥ BC.

Answer: AD/DB = AE/EC = 1/3, so DE ∥ BC

1. Slope = (y₂ − y₁)/(x₂ − x₁) = (−6 − 10)/(14 − 14) = −16/0.
2. Division by zero: both points have x = 14, so the line is vertical.

Answer: The slope is undefined (the line x = 14 is vertical)

1. Intercept form: x/a + y/b = 1 with a = 4, b = −6: x/4 − y/6 = 1.
2. Multiply by 12: 3x − 2y = 12.

Answer: 3x − 2y − 12 = 0

1. Multiply the top and bottom inside the root by (1 + cos θ): √[(1 + cos θ)² / (1 − cos²θ)].
2. 1 − cos²θ = sin²θ, so this is √[(1 + cos θ)² / sin²θ] = (1 + cos θ)/sin θ (taking sin θ > 0, as the identity is meant for).
3. (1 + cos θ)/sin θ = 1/sin θ + cos θ/sin θ = cosec θ + cot θ.

Answer: Left side = cosec θ + cot θ — proved

1. For the first n natural numbers, σ = √[(n² − 1)/12].
2. n = 21: σ = √[(441 − 1)/12] = √(440/12) = √36.67.

Answer: σ ≈ 6.06

1. The sample space has 6 × 2 = 12 outcomes.
2. Favourable: (1, H), (3, H), (5, H) — 3 outcomes.
3. P = 3/12.

Answer: 1/4

1. Let the cube's side be a. The largest sphere inside touches all faces, so its diameter is a and radius a/2.
2. Volume of cube = a³; volume of sphere = (4/3)π(a/2)³ = πa³/6.
3. Ratio = a³ : πa³/6 = 6 : π.

Answer: 6 : π (about 21 : 11 with π = 22/7)

1. A = {0, 1}, B = {2, 3, 4}, C = {3, 5}.
2. B ∩ C = {3}, so A × (B ∩ C) = {(0, 3), (1, 3)}.
3. A × B = {(0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (1, 4)}; A × C = {(0, 3), (0, 5), (1, 3), (1, 5)}.
4. (A × B) ∩ (A × C) = {(0, 3), (1, 3)} — the same set.

Answer: Both sides = {(0, 3), (1, 3)} — verified

1. f∘g(x) = f(3x + 1) = 3x + 1 − 1 = 3x, so ((f∘g)∘h)(x) = f∘g(x²) = 3x².
2. g∘h(x) = g(x²) = 3x² + 1, so (f∘(g∘h))(x) = f(3x² + 1) = 3x² + 1 − 1 = 3x².
3. Both are 3x² for every x.

Answer: (f∘g)∘h = f∘(g∘h) = 3x² — proved

1. The first is 301 (7 × 43) and the last is 595 (7 × 85): an A.P. with a = 301, l = 595, d = 7.
2. n = (595 − 301)/7 + 1 = 42 + 1 = 43.
3. S = n/2 (a + l) = 43/2 × 896 = 43 × 448.

Answer: 19264

1. Sₙ = 3(1 + 11 + 111 + …) = (3/9)(9 + 99 + 999 + …) = (1/3)[(10 − 1) + (100 − 1) + (1000 − 1) + …].
2. = (1/3)[(10 + 100 + … + 10ⁿ) − n].
3. 10 + 100 + … + 10ⁿ is a G.P.: 10(10ⁿ − 1)/9.
4. Sₙ = (1/3)[10(10ⁿ − 1)/9 − n] = 10(10ⁿ − 1)/27 − n/3.

Answer: Sₙ = 10(10ⁿ − 1)/27 − n/3

1. AB (2 × 2): row 1: 5 + 2 + 45 = 52 and 35 + 4 − 9 = 30; row 2: 1 + 2 + 40 = 43 and 7 + 4 − 8 = 3. AB = rows (52 30), (43 3).
2. (AB)ᵀ = rows (52 43), (30 3).
3. Bᵀ = rows (1 1 5), (7 2 −1); Aᵀ = rows (5 1), (2 2), (9 8).
4. BᵀAᵀ: row 1: 5 + 2 + 45 = 52 and 1 + 2 + 40 = 43; row 2: 35 + 4 − 9 = 30 and 7 + 4 − 8 = 3. BᵀAᵀ = rows (52 43), (30 3).

Answer: (AB)ᵀ = BᵀAᵀ = rows (52 43), (30 3) — verified

1. In ΔABC (sides a = BC, b = CA, c = AB) let the bisectors of A, B and C meet BC, CA and AB at D, E and F.
2. By the angle bisector theorem: BD/DC = AB/AC = c/b, CE/EA = BC/BA = a/c, AF/FB = CA/CB = b/a.
3. Multiply: (BD/DC)(CE/EA)(AF/FB) = (c/b)(a/c)(b/a) = 1.
4. By the converse of Ceva's theorem, AD, BE and CF meet at one point.

Answer: The three angle bisectors meet at one point (the incentre) — shown

1. Take the vertices in order (anticlockwise): A(8, 6), B(5, 11), C(−5, 12), D(−4, 3).
2. Area = ½ |(x₁y₂ + x₂y₃ + x₃y₄ + x₄y₁) − (y₁x₂ + y₂x₃ + y₃x₄ + y₄x₁)|.
3. x₁y₂ + … = 88 + 60 − 15 − 24 = 109; y₁x₂ + … = 30 − 55 − 48 + 24 = −49.
4. Area = ½ |109 − (−49)| = ½ × 158.

Answer: 79 square units

1. Slope of BC = (3 − (−2))/(12 − 10) = 5/2.
2. The altitude from A is perpendicular to BC: slope = −2/5.
3. Through A(−3, 0): y − 0 = −2/5 (x + 3) → 5y = −2x − 6.

Answer: 2x + 5y + 6 = 0

1. Ship at 30°: distance = 200 / tan 30° = 200√3 m.
2. Ship at 45°: distance = 200 / tan 45° = 200 m.
3. They are on opposite sides, so the distance between them = 200√3 + 200 = 200(1.732 + 1) = 200 × 2.732.

Answer: 546.4 m

1. One revolution covers the curved surface area: 2πrh with r = 1.4 m, h = 3 m.
2. 2 × 22/7 × 1.4 × 3 = 26.4 m².
3. 8 revolutions: 8 × 26.4 = 211.2 m².

Answer: 211.2 m²

1. Cone: r = 7, h = 8, volume = (1/3)π × 49 × 8 = 392π/3 cm³.
2. Hollow sphere: external radius R = 5, internal radius r: volume = (4/3)π(R³ − r³) = (4/3)π(125 − r³).
3. (4/3)π(125 − r³) = 392π/3 → 4(125 − r³) = 392 → 125 − r³ = 98 → r³ = 27 → r = 3 cm.

Answer: Internal diameter = 6 cm

1. Mean = (24 + 26 + 33 + 37 + 29 + 31)/6 = 180/6 = 30.
2. Deviations: −6, −4, 3, 7, −1, 1; squares: 36, 16, 9, 49, 1, 1; sum = 112.
3. σ = √(112/6) = √18.67 ≈ 4.32.
4. C.V. = σ/mean × 100 = 4.32/30 × 100.

Answer: C.V. ≈ 14.40%

1. n(S) = 36.
2. A (doublet) = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}: n(A) = 6. B (sum 4) = {(1,3), (2,2), (3,1)}: n(B) = 3. A ∩ B = {(2,2)}: 1.
3. P(A ∪ B) = 6/36 + 3/36 − 1/36 = 8/36.

Answer: 2/9

1. α + β = 3/2 and αβ = 1/2.
2. α² + β² = (α + β)² − 2αβ = 9/4 − 1 = 5/4.
3. Sum of new roots = 1/α² + 1/β² = (α² + β²)/(αβ)² = (5/4)/(1/4) = 5. Product = 1/(αβ)² = 4.
4. Equation: x² − (sum)x + product = 0.

Answer: x² − 5x + 4 = 0

1. Draw any triangle PQR (the question leaves its size to you).
2. From Q draw a ray QX making an acute angle with QR on the side opposite P.
3. Mark 5 equal arcs Q₁, Q₂, …, Q₅ along QX (5 = the larger of 3 and 5).
4. Join Q₅R. Through Q₃ draw a line parallel to Q₅R, meeting QR at R′.
5. Through R′ draw a line parallel to RP, meeting QP at P′.
6. ΔP′QR′ is the required triangle: each side is 3/5 of the matching side of ΔPQR.

Answer: ΔP′QR′ with QR′ = 3/5 QR (steps above)

1. Draw a circle with centre O and radius 3 cm. Mark P with OP = 8 cm.
2. Bisect OP; with its midpoint M as centre and MO as radius, draw a circle cutting the first circle at A and B.
3. Join PA and PB: these are the tangents (∠OAP = ∠OBP = 90°, angles in a semicircle).
4. Check by calculation: PA = √(OP² − OA²) = √(64 − 9) = √55 ≈ 7.4 cm.

Answer: PA = PB ≈ 7.4 cm

1. Table: x = 0, 1, 2, 3, 4, 5, 6 gives y = 9, 4, 1, 0, 1, 4, 9.
2. Plot the points and join them with a smooth curve (a parabola opening upwards).
3. The curve touches the X axis at only one point, (3, 0).
4. So the equation has real and equal roots: x = 3, 3 (check: discriminant 36 − 36 = 0).

Answer: Real and equal roots: x = 3, 3

1. Table: x = 2, 4, 6, 8, 10 gives y = 1, 2, 3, 4, 5. Plot and join: a straight line through the origin.
2. y = kx with k = ½: the constant of variation is ½. On the graph, y/x = 1/2 at every plotted point.
3. (i) x = 9: y = 9/2 = 4.5. (ii) y = 7.5: x = 7.5 × 2 = 15.

Answer: k = 1/2; y = 4.5 when x = 9; x = 15 when y = 7.5