Maharashtra SSC Maths Model Paper 2027 — Full Paper With Solutions

Maharashtra SSC Maths has two separate papers — Part 1 (Algebra) and Part 2 (Geometry), each carrying 40 marks. Total: 80 marks in 2 hours each.

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PART 1 — ALGEBRA (40 Marks)

Paper Structure

Section	Type	Marks
Section A	MCQ (4 questions)	8
Section B	Short Answer 1 (2 marks)	10
Section C	Short Answer 2 (3 marks)	9
Section D	Long Answer (4 marks)	12

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SECTION A — MCQ (2 Marks Each)

Q1. The discriminant of the quadratic equation 2x² − 5x + 3 = 0 is:

(a) 1 (b) −1 (c) 24 (d) −24

Answer: (a) 1
(b² − 4ac = 25 − 4×2×3 = 25 − 24 = 1)

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Q2. The sum of the first 20 natural numbers is:

(a) 190 (b) 200 (c) 210 (d) 220

Answer: (c) 210
(Sₙ = n(n+1)/2 = 20×21/2 = 210)

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Q3. If two coins are tossed simultaneously, the probability of getting exactly one head is:

(a) 1/4 (b) 1/2 (c) 3/4 (d) 1

Answer: (b) 1/2
(Favourable outcomes: HT, TH = 2; Total = 4; P = 2/4 = 1/2)

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Q4. The value of the determinant |[3, 2], [1, 4]| is:

(a) 10 (b) 14 (c) 12 (d) 8

Answer: (a) 10
(3×4 − 2×1 = 12 − 2 = 10)

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SECTION B — Short Answer 1 (2 Marks Each)

Q5. Find the roots of x² − 5x + 6 = 0 by factorisation.

Solution:
x² − 5x + 6 = 0
x² − 3x − 2x + 6 = 0
x(x−3) − 2(x−3) = 0
(x−2)(x−3) = 0
x = 2 or x = 3

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Q6. Find the 15th term of the AP: 3, 7, 11, 15, ...

Solution:
a = 3, d = 4, n = 15
aₙ = a + (n−1)d = 3 + 14×4 = 3 + 56 = 59

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Q7. A card is drawn at random from a deck of 52 cards. Find the probability that it is a king.

Solution:
Number of kings = 4
Total cards = 52
P(king) = 4/52 = 1/13

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Q8. Solve using Cramer's rule: x + y = 5, x − y = 1.

Solution:
D = |[1,1],[1,-1]| = −1−1 = −2
Dx = |[5,1],[1,-1]| = −5−1 = −6
Dy = |[1,5],[1,1]| = 1−5 = −4
x = Dx/D = −6/−2 = 3; y = Dy/D = −4/−2 = 2

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SECTION C — Short Answer 2 (3 Marks Each)

Q9. If α and β are roots of 3x² − 5x + 2 = 0, find: (i) α + β (ii) αβ (iii) α² + β²

Solution:
α + β = 5/3; αβ = 2/3
α² + β² = (α+β)² − 2αβ = (5/3)² − 2(2/3) = 25/9 − 4/3 = 25/9 − 12/9 = 13/9

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Q10. The sum of three consecutive terms of an AP is 27, and their product is 504. Find the terms.

Solution:
Let terms be a−d, a, a+d
Sum: 3a = 27 → a = 9
Product: (9−d)(9)(9+d) = 504
9(81−d²) = 504
81−d² = 56
d² = 25 → d = 5
Terms: 4, 9, 14

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Q11. Find the mean of the following data:

Class	0–10	10–20	20–30	30–40	40–50
Freq	5	8	15	10	2

Solution:
midpoints: 5, 15, 25, 35, 45
Σfx = 5×5 + 8×15 + 15×25 + 10×35 + 2×45 = 25+120+375+350+90 = 960
Σf = 40
Mean = 960/40 = 24

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SECTION D — Long Answer (4 Marks Each)

Q12. The sum of a two-digit number and the number formed by interchanging its digits is 110. The difference of the digits is 2. Find the number.

Solution:
Let tens digit = x, units digit = y
Number = 10x + y; Reversed = 10y + x
(10x+y) + (10y+x) = 110 → 11(x+y) = 110 → x+y = 10 ...(i)
x − y = 2 ...(ii)
Adding: 2x = 12 → x = 6, y = 4
Number = 64 (or 46 both satisfy the conditions)

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PART 2 — GEOMETRY (40 Marks)

SECTION A — MCQ (2 Marks Each)

Q13. In △ABC, if DE ∥ BC and AD/DB = 3/5, then AE/EC is:

(a) 5/3 (b) 3/5 (c) 3/8 (d) 5/8

Answer: (b) 3/5
(By Basic Proportionality Theorem, AD/DB = AE/EC = 3/5)

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Q14. The length of tangent from an external point P to a circle of radius 5 cm, if OP = 13 cm, is:

(a) 8 cm (b) 10 cm (c) 12 cm (d) 18 cm

Answer: (c) 12 cm
(PT² = OP² − r² = 169 − 25 = 144; PT = 12 cm)

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SECTION B — Short Answer (2 Marks Each)

Q15. In a right triangle, the hypotenuse is 13 cm and one leg is 5 cm. Find the other leg.

Solution:
By Pythagoras: a² + b² = c²
5² + b² = 13²
b² = 169 − 25 = 144
b = 12 cm

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Q16. Find the area of a sector with radius 7 cm and angle 60°.

Solution:
Area = (θ/360) × πr² = (60/360) × (22/7) × 7²
= (1/6) × 22 × 7 = 154/6 = 25.67 cm²

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SECTION C — Short Answer 2 (3 Marks Each)

Q17. Prove that tangents drawn from an external point to a circle are equal in length.

Proof:
Let PA and PB be two tangents from external point P to circle with centre O.
OA ⊥ PA and OB ⊥ PB (radius ⊥ tangent)
In △OAP and △OBP:
OA = OB (radii), OP = OP (common), ∠OAP = ∠OBP = 90°
By RHS congruence: △OAP ≅ △OBP
Therefore PA = PB ∎

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SECTION D — Long Answer (4 Marks Each)

Q18. A 15 m long ladder reaches a window 12 m above the ground. Find how far the foot of the ladder is from the wall. Also find the angle the ladder makes with the ground.

Solution:
By Pythagoras: d² = 15² − 12² = 225 − 144 = 81
d = 9 m from wall
sin θ = 12/15 = 4/5 = 0.8
θ = sin⁻¹(0.8) ≈ 53°8'

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Practice Problems (Self Test)

1. Find the quadratic equation whose roots are 3 and −5.
2. How many terms of AP 9, 17, 25... must be taken to give sum 636?
3. Find the circumference and area of a circle with radius 14 cm.
4. Prove: In a right triangle, square of hypotenuse = sum of squares of other two sides.

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Chapter-Wise Weightage

Chapter	Marks
Quadratic Equations	8
Arithmetic Progression	8
Probability and Statistics	8
Linear Equations (Algebra)	8
Triangles and Similarity	8
Circles and Tangents	8
Mensuration	8
Coordinate Geometry	8

Tip: Maharashtra SSC awards full marks for correct method even if the final answer has a small arithmetic error. Always show your complete working.