AP SSC Maths Model Paper 2027 — Full Paper With Solutions

This model paper follows the official BSEAP pattern for AP SSC Mathematics 2027. Total: 100 marks in 3 hours 15 minutes.

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Paper Structure

Section	Type	Marks
Part A	MCQ — 20 questions	20
Part B	Fill in the blanks — 10	10
Part C	Short Answer — 7 of 10 (4 marks each)	28
Part D	Long Answer — 4 of 6 (6 marks each)	24
Internal	Project Work	18

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PART A — MCQ (1 Mark Each)

Q1. The value of sin 30° × cos 60° is:

(a) 1 (b) 1/2 (c) 1/4 (d) √3/2

Answer: (c) 1/4
(sin 30° = 1/2; cos 60° = 1/2; product = 1/4)

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Q2. If one zero of the polynomial p(x) = x² − 4x + k is 2, then k is:

(a) 2 (b) 4 (c) 6 (d) 8

Answer: (b) 4
(p(2) = 4 − 8 + k = 0 → k = 4)

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Q3. The sum of first n natural numbers is:

(a) n(n+1) (b) n(n+1)/2 (c) n² (d) n(n−1)/2

Answer: (b) n(n+1)/2

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Q4. The distance of point (3, 4) from the origin is:

(a) 3 (b) 4 (c) 5 (d) 7

Answer: (c) 5
(d = √(9+16) = 5)

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Q5. Which of the following is NOT a quadratic equation?

(a) x² + 2x + 1 = 0 (b) x³ − x = 0 (c) x² = 4 (d) 2x² − 3x + 1 = 0

Answer: (b) x³ − x = 0
(Degree 3 — cubic, not quadratic)

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Q6. Volume of a cylinder of radius r and height h is:

(a) πr²h (b) 2πrh (c) πr²h/3 (d) 4πr²

Answer: (a) πr²h

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Q7. The mean of 5, 10, 15, 20, 25 is:

(a) 10 (b) 12 (c) 15 (d) 20

Answer: (c) 15
(Sum = 75; Mean = 75/5 = 15)

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Q8. The probability of getting a prime number when a die is thrown is:

(a) 1/6 (b) 1/3 (c) 1/2 (d) 2/3

Answer: (c) 1/2
(Prime numbers on die: 2, 3, 5 = 3 outcomes; P = 3/6 = 1/2)

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Q9. In △ABC, if DE ∥ BC, AD = 2 cm, DB = 3 cm, then AE/EC = ?

(a) 2/3 (b) 3/2 (c) 2/5 (d) 3/5

Answer: (a) 2/3
(By BPT: AD/DB = AE/EC = 2/3)

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Q10. A tangent to a circle makes an angle of _______ with the radius at the point of tangency.

(a) 45° (b) 60° (c) 90° (d) 180°

Answer: (c) 90°

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PART B — Fill in the Blanks (1 Mark Each)

Q11. The sum of zeroes of x² − 5x + 6 is ________. Answer: 5

Q12. If 2 is a root of x² + kx − 4 = 0, then k = ________. Answer: 0
(4 + 2k − 4 = 0 → k = 0)

Q13. The nth term of AP a, a+d, a+2d... is ________. Answer: a + (n−1)d

Q14. sin²θ + cos²θ = ________. Answer: 1

Q15. The area of a circle with radius r is ________. Answer: πr²

Q16. Median of 3, 7, 9, 11, 15 is ________. Answer: 9

Q17. If two triangles are similar, their corresponding angles are ________. Answer: Equal

Q18. LCM × HCF = ________ for two numbers a and b. Answer: a × b

Q19. The graph of a quadratic polynomial is a ________. Answer: Parabola

Q20. A circle has ________ tangents from an external point. Answer: Two

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PART C — Short Answer (4 Marks Each — Attempt 7 of 10)

Q21. Find the zeroes of polynomial p(x) = 2x² − 5x + 3 and verify the relationship between zeroes and coefficients.

Solution:
2x² − 5x + 3 = 0
2x² − 2x − 3x + 3 = 0
2x(x−1) − 3(x−1) = 0
(2x−3)(x−1) = 0
Zeroes: α = 3/2, β = 1

Verification:
α + β = 3/2 + 1 = 5/2 = −(−5)/2 = −b/a ✓
αβ = 3/2 × 1 = 3/2 = c/a ✓

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Q22. Solve graphically: x + y = 6 and x − y = 2.

Solution:
For x + y = 6: when x=0, y=6; when x=6, y=0; when x=3, y=3
For x − y = 2: when x=0, y=−2; when x=2, y=0; when x=4, y=2

Graph: Plot both lines. They intersect at point (4, 2).

Verification: 4+2=6 ✓ and 4−2=2 ✓

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Q23. A tower stands vertically on the ground. From a point on the ground 15 m away from the base, the angle of elevation of the top is 60°. Find the height of the tower.

Solution:
Let height of tower = h
tan 60° = h/15
√3 = h/15
h = 15√3 ≈ 25.98 m

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Q24. Find the area of a sector with radius 14 cm and central angle 90°. Also find the arc length.

Solution:
Area of sector = (θ/360°) × πr²
= (90/360) × (22/7) × 196
= (1/4) × 616 = 154 cm²

Arc length = (θ/360°) × 2πr
= (90/360) × 2 × (22/7) × 14
= (1/4) × 88 = 22 cm

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PART D — Long Answer (6 Marks Each — Attempt 4 of 6)

Q25. The following table shows the marks obtained by 100 students. Find the median.

Marks	0–10	10–20	20–30	30–40	40–50
Students	5	15	30	35	15

Solution:

Marks	Freq	Cumulative Freq
0–10	5	5
10–20	15	20
20–30	30	50
30–40	35	85
40–50	15	100

n/2 = 50 → Median class = 20–30
l = 20, f = 30, cf = 20, h = 10

Median = l + [(n/2 − cf)/f] × h
= 20 + [(50−20)/30] × 10
= 20 + (30/30) × 10
= 30

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Q26. Prove: The tangent at any point of a circle is perpendicular to the radius through the point of contact.

Solution:
Given: Circle with centre O; P is point of contact; TP is tangent at P.
To Prove: OP ⊥ TP

Proof:
Let Q be any other point on tangent TP.
Since tangent touches circle at P only, Q lies outside the circle.
OQ = OP + PQ > OP (as OQ passes outside)

So OP is the shortest distance from O to line TP.
The shortest distance from a point to a line is the perpendicular distance.
Therefore OP ⊥ TP ∎

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Practice Problems (Self Test)

1. Find three consecutive terms of AP whose sum is 27 and product is 504.
2. Prove: √5 is irrational.
3. A cone of height 24 cm has a base radius of 7 cm. Find its slant height, CSA and volume.
4. From a bag containing 5 red, 3 blue, 2 green balls, find P(not red).

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Chapter-Wise Weightage

Chapter	Marks
Real Numbers and Polynomials	10
Pair of Linear Equations	8
Quadratic Equations	8
Arithmetic Progressions	8
Triangles and Similarity	10
Trigonometry	10
Coordinate Geometry	8
Circles and Constructions	8
Mensuration	10
Statistics and Probability	10
Internal/Project	10

Tip: AP SSC Maths Part B (Fill in the Blanks) is 10 free marks — purely formula and definition based. Cover all basic formulas: area, volume, trigonometric ratios, AP formulas, and distance formula. These 10 marks require zero complex solving.