BOARD OF SECONDARY EDUCATION, TELANGANA
TS SSC CLASS 10 MATHEMATICS PUBLIC EXAMINATION 2022
Official Question Paper & Step-by-Step Solutions | Muqeem Tutorial
PART - A (60 MARKS) — SUBJECTIVE QUESTIONS
SECTION - I (6 × 2 = 12 Marks) — Very Short Answer Questions (Answer ALL)
Q1. Use Euclid's division algorithm to find the HCF of 306 and 657.
Solution:
1. 657 = 306 × 2 + 45
2. 306 = 45 × 6 + 36
3. 45 = 36 × 1 + 9
4. 36 = 9 × 4 + 0
👉 HCF(306, 657) = 9
1. 657 = 306 × 2 + 45
2. 306 = 45 × 6 + 36
3. 45 = 36 × 1 + 9
4. 36 = 9 × 4 + 0
👉 HCF(306, 657) = 9
Q2. If A = {x : x ∈ N, x < 6} and B = {x : x is a prime factor of 30}, find A ∩ B.
Solution:
• A = {1, 2, 3, 4, 5}
• B = {2, 3, 5}
👉 A ∩ B = {2, 3, 5}
• A = {1, 2, 3, 4, 5}
• B = {2, 3, 5}
👉 A ∩ B = {2, 3, 5}
Q3. Find a quadratic polynomial whose sum of zeroes is -3 and product of zeroes is 2.
Solution:
• Formula: x2 - (Sum)x + Product
• x2 - (-3)x + 2 = x2 + 3x + 2
👉 Required Polynomial = x2 + 3x + 2
• Formula: x2 - (Sum)x + Product
• x2 - (-3)x + 2 = x2 + 3x + 2
👉 Required Polynomial = x2 + 3x + 2
Q4. Find the sum of the first 20 terms of the AP: 1, 4, 7, 10...
Solution:
• a = 1, d = 3, n = 20
• S20 = (20/2)[2(1) + (19)(3)] = 10[2 + 57] = 10 × 59 = 590
👉 Sum S20 = 590
• a = 1, d = 3, n = 20
• S20 = (20/2)[2(1) + (19)(3)] = 10[2 + 57] = 10 × 59 = 590
👉 Sum S20 = 590
Q5. Find the area of the triangle with vertices (2, 3), (-1, 0), and (2, -4).
Solution:
• Area = (1/2) | x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2) |
• Area = (1/2) | 2(0 - (-4)) + (-1)(-4 - 3) + 2(3 - 0) |
• Area = (1/2) | 8 + 7 + 6 | = 21/2 = 10.5
👉 Area = 10.5 sq. units
• Area = (1/2) | x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2) |
• Area = (1/2) | 2(0 - (-4)) + (-1)(-4 - 3) + 2(3 - 0) |
• Area = (1/2) | 8 + 7 + 6 | = 21/2 = 10.5
👉 Area = 10.5 sq. units
Q6. If 15 cot A = 8, find sin A and sec A.
Solution:
• cot A = 8/15 ⇒ Adjacent = 8, Opposite = 15
• Hypotenuse = √(82 + 152) = √(64 + 225) = √289 = 17
👉 sin A = 15/17, sec A = 17/8
• cot A = 8/15 ⇒ Adjacent = 8, Opposite = 15
• Hypotenuse = √(82 + 152) = √(64 + 225) = √289 = 17
👉 sin A = 15/17, sec A = 17/8
SECTION - II (6 × 4 = 24 Marks) — Short Answer Questions (Answer ALL)
Q7. Prove that 3 + 2√5 is an irrational number.
Solution:
1. Assume 3 + 2√5 = a/b (a, b coprime integers, b ≠ 0).
2. 2√5 = (a/b) - 3 ⇒ √5 = (a - 3b) / (2b).
3. Since a, b are integers, (a - 3b)/(2b) is rational, implying √5 is rational.
4. Contradiction! √5 is irrational.
👉 Hence, 3 + 2√5 is Irrational. (Proved)
1. Assume 3 + 2√5 = a/b (a, b coprime integers, b ≠ 0).
2. 2√5 = (a/b) - 3 ⇒ √5 = (a - 3b) / (2b).
3. Since a, b are integers, (a - 3b)/(2b) is rational, implying √5 is rational.
4. Contradiction! √5 is irrational.
👉 Hence, 3 + 2√5 is Irrational. (Proved)
Q8. Find the roots of the quadratic equation x2 - 3x - 10 = 0 by using the quadratic formula.
Solution:
• x = [-b ± √(b2 - 4ac)] / 2a
• x = [-(-3) ± √((-3)2 - 4(1)(-10))] / 2 = [3 ± √(9 + 40)] / 2 = [3 ± 7] / 2
• x = 10/2 = 5 or x = -4/2 = -2
👉 Roots are x = 5 and x = -2
• x = [-b ± √(b2 - 4ac)] / 2a
• x = [-(-3) ± √((-3)2 - 4(1)(-10))] / 2 = [3 ± √(9 + 40)] / 2 = [3 ± 7] / 2
• x = 10/2 = 5 or x = -4/2 = -2
👉 Roots are x = 5 and x = -2
Q9. Solve the system of linear equations by elimination method: 3x + 4y = 10 and 2x - 2y = 2.
Solution:
• Multiply 2nd equation by 2: 4x - 4y = 4
• Add to 1st equation: (3x + 4y) + (4x - 4y) = 10 + 4 ⇒ 7x = 14 ⇒ x = 2
• Substitute x = 2 in 2(2) - 2y = 2 ⇒ 4 - 2y = 2 ⇒ 2y = 2 ⇒ y = 1
👉 Solution: x = 2, y = 1
• Multiply 2nd equation by 2: 4x - 4y = 4
• Add to 1st equation: (3x + 4y) + (4x - 4y) = 10 + 4 ⇒ 7x = 14 ⇒ x = 2
• Substitute x = 2 in 2(2) - 2y = 2 ⇒ 4 - 2y = 2 ⇒ 2y = 2 ⇒ y = 1
👉 Solution: x = 2, y = 1
Q10. Find a point on the X-axis which is equidistant from A(2, -5) and B(-2, 9).
Solution:
• Point on X-axis P = (x, 0). Given PA2 = PB2
• (x - 2)2 + (0 - (-5))2 = (x - (-2))2 + (0 - 9)2
• x2 - 4x + 4 + 25 = x2 + 4x + 4 + 81
• -4x + 29 = 4x + 85 ⇒ 8x = -56 ⇒ x = -7
👉 Required Point P = (-7, 0)
• Point on X-axis P = (x, 0). Given PA2 = PB2
• (x - 2)2 + (0 - (-5))2 = (x - (-2))2 + (0 - 9)2
• x2 - 4x + 4 + 25 = x2 + 4x + 4 + 81
• -4x + 29 = 4x + 85 ⇒ 8x = -56 ⇒ x = -7
👉 Required Point P = (-7, 0)
Q11. A box contains 5 red marbles, 8 white marbles, and 4 green marbles. If one marble is drawn at random, find the probability that it is: (i) red, (ii) white, (iii) not green.
Solution:
• Total marbles n(S) = 5 + 8 + 4 = 17
(i) P(red) = 5/17
(ii) P(white) = 8/17
(iii) P(not green) = (17 - 4) / 17 = 13/17
👉 Probabilities: (i) 5/17, (ii) 8/17, (iii) 13/17
• Total marbles n(S) = 5 + 8 + 4 = 17
(i) P(red) = 5/17
(ii) P(white) = 8/17
(iii) P(not green) = (17 - 4) / 17 = 13/17
👉 Probabilities: (i) 5/17, (ii) 8/17, (iii) 13/17
Q12. Prove that (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A.
Solution:
• Expanding LHS = sin2 A + cosec2 A + 2 + cos2 A + sec2 A + 2
• Since sin2 A + cos2 A = 1, LHS = 1 + 4 + (1 + cot2 A) + (1 + tan2 A)
• LHS = 7 + tan2 A + cot2 A = RHS
👉 Hence Proved.
• Expanding LHS = sin2 A + cosec2 A + 2 + cos2 A + sec2 A + 2
• Since sin2 A + cos2 A = 1, LHS = 1 + 4 + (1 + cot2 A) + (1 + tan2 A)
• LHS = 7 + tan2 A + cot2 A = RHS
👉 Hence Proved.
SECTION - III (4 × 6 = 24 Marks) — Essay Questions with Internal Choice
Q13 (A). Solve the pair of linear equations graphically: x + y = 5 and 2x - y = 4.
Solution (13A):
• Line 1 (x + y = 5): Points (0, 5) and (5, 0)
• Line 2 (2x - y = 4): Points (0, -4) and (2, 0)
• Intersecting point on graph = (3, 2).
👉 Solution: x = 3, y = 2
• Line 1 (x + y = 5): Points (0, 5) and (5, 0)
• Line 2 (2x - y = 4): Points (0, -4) and (2, 0)
• Intersecting point on graph = (3, 2).
👉 Solution: x = 3, y = 2
--- OR ---
Q13 (B). Draw the graph of polynomial p(x) = x2 - x - 6 and find its zeroes.
Solution (13B):
• Factorizing: (x - 3)(x + 2) = 0. Parabola intersects X-axis at (-2, 0) and (3, 0).
👉 Zeroes are x = 3 and x = -2
• Factorizing: (x - 3)(x + 2) = 0. Parabola intersects X-axis at (-2, 0) and (3, 0).
👉 Zeroes are x = 3 and x = -2
Q14 (A). State and prove Converse of Basic Proportionality Theorem.
Solution (14A):
• Statement: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
• Proof: In ΔABC, let DE divide AB and AC such that AD/DB = AE/EC. Assume DE is not parallel to BC and draw DE' || BC. Then AD/DB = AE'/E'C. Therefore AE/EC = AE'/E'C, which implies E and E' coincide.
👉 DE || BC. (Proved)
• Statement: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
• Proof: In ΔABC, let DE divide AB and AC such that AD/DB = AE/EC. Assume DE is not parallel to BC and draw DE' || BC. Then AD/DB = AE'/E'C. Therefore AE/EC = AE'/E'C, which implies E and E' coincide.
👉 DE || BC. (Proved)
--- OR ---
Q14 (B). A metallic sphere of radius 4.2 cm is melted and recast into a cylinder of radius 6 cm. Find the height of the cylinder.
Solution (14B):
• Volume of Sphere = (4/3)πr3 = (4/3) × π × (4.2)3 = 98.784π cm3
• Volume of Cylinder = πR2h = π × 62 × h = 36πh
• Equating volumes: 36πh = 98.784π ⇒ h = 2.74 cm
👉 Height of Cylinder = 2.74 cm
• Volume of Sphere = (4/3)πr3 = (4/3) × π × (4.2)3 = 98.784π cm3
• Volume of Cylinder = πR2h = π × 62 × h = 36πh
• Equating volumes: 36πh = 98.784π ⇒ h = 2.74 cm
👉 Height of Cylinder = 2.74 cm
Q15 (A). The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.
Solution (15A):
• Let tower height = h. Angles are θ and (90° - θ).
• tan θ = h / 4
• tan (90° - θ) = cot θ = h / 9 ⇒ tan θ = 9 / h
• Equating: h / 4 = 9 / h ⇒ h2 = 36 ⇒ h = 6 m.
👉 Height of Tower = 6 m. (Proved)
• Let tower height = h. Angles are θ and (90° - θ).
• tan θ = h / 4
• tan (90° - θ) = cot θ = h / 9 ⇒ tan θ = 9 / h
• Equating: h / 4 = 9 / h ⇒ h2 = 36 ⇒ h = 6 m.
👉 Height of Tower = 6 m. (Proved)
--- OR ---
Q15 (B). Prove that the angle between two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the center.
Solution (15B):
• Quadrilateral OAPB has radii OA ⊥ AP and OB ⊥ BP ⇒ ∠OAP = 90° and ∠OBP = 90°.
• Sum of angles of quadrilateral = 360° ⇒ ∠APB + ∠AOB + 90° + 90° = 360°
• ∠APB + ∠AOB = 180° (Supplementary).
👉 Hence Proved.
• Quadrilateral OAPB has radii OA ⊥ AP and OB ⊥ BP ⇒ ∠OAP = 90° and ∠OBP = 90°.
• Sum of angles of quadrilateral = 360° ⇒ ∠APB + ∠AOB + 90° + 90° = 360°
• ∠APB + ∠AOB = 180° (Supplementary).
👉 Hence Proved.
Q16 (A). Find the Median of the following distribution:
| Marks | 0 - 10 | 10 - 20 | 20 - 30 | 30 - 40 | 40 - 50 |
|---|---|---|---|---|---|
| Students | 5 | 15 | 20 | 10 | 10 |
Solution (16A):
• Cumulative Frequencies (cf) = 5, 20, 40, 50, 60. Total N = 60 ⇒ N/2 = 30.
• Median Class = 20-30 (L = 20, cf = 20, f = 20, h = 10)
• Median = L + [(N/2 - cf)/f] × h = 20 + [(30 - 20)/20] × 10 = 20 + 5 = 25
👉 Median = 25
• Cumulative Frequencies (cf) = 5, 20, 40, 50, 60. Total N = 60 ⇒ N/2 = 30.
• Median Class = 20-30 (L = 20, cf = 20, f = 20, h = 10)
• Median = L + [(N/2 - cf)/f] × h = 20 + [(30 - 20)/20] × 10 = 20 + 5 = 25
👉 Median = 25
--- OR ---
Q16 (B). Two dice are thrown simultaneously. Find the probability of getting: (i) a doublet, (ii) sum = 10, (iii) sum less than 5.
Solution (16B):
• Total outcomes n(S) = 36
(i) Doublet = 6/36 = 1/6
(ii) Sum = 10: {(4,6),(5,5),(6,4)} ⇒ 3 outcomes ⇒ P = 3/36 = 1/12
(iii) Sum < 5: {(1,1),(1,2),(1,3),(2,1),(2,2),(3,1)} ⇒ 6 outcomes ⇒ P = 6/36 = 1/6
👉 Probabilities: (i) 1/6, (ii) 1/12, (iii) 1/6
• Total outcomes n(S) = 36
(i) Doublet = 6/36 = 1/6
(ii) Sum = 10: {(4,6),(5,5),(6,4)} ⇒ 3 outcomes ⇒ P = 3/36 = 1/12
(iii) Sum < 5: {(1,1),(1,2),(1,3),(2,1),(2,2),(3,1)} ⇒ 6 outcomes ⇒ P = 6/36 = 1/6
👉 Probabilities: (i) 1/6, (ii) 1/12, (iii) 1/6
PART - B (20 MARKS) — MULTIPLE CHOICE QUESTIONS (1 Mark Each)
1. The discriminant of quadratic equation 3x2 - 2x + 1/3 = 0 is:
(A) 0 (B) 1 (C) 2 (D) -1
👉 Correct Answer: (A) 0 [(-2)2 - 4(3)(1/3) = 4 - 4 = 0]
2. Distance of point (-6, 8) from the origin is:
(A) 8 (B) -6 (C) 10 (D) 14
👉 Correct Answer: (C) 10 [√(36 + 64) = 10]
3. The value of sin2 60° + cos2 60° is:
(A) 0 (B) 1 (C) 2 (D) 1/2
👉 Correct Answer: (B) 1
4. If P(E) = 0.26, then P(NOT E) is:
(A) 0.74 (B) 0.26 (C) 1.26 (D) 0.50
👉 Correct Answer: (A) 0.74 [1 - 0.26 = 0.74]
5. The centroid of a triangle divides each median in the ratio:
(A) 1 : 1 (B) 2 : 1 (C) 3 : 1 (D) 1 : 2
👉 Correct Answer: (B) 2 : 1
6. Mid-point of class interval 25 - 35 is:
(A) 25 (B) 30 (C) 35 (D) 60
👉 Correct Answer: (B) 30 [(25+35)/2 = 30]
7. Next term of AP: √2, √8, √18, √32... is:
(A) √50 (B) √60 (C) √40 (D) √48
👉 Correct Answer: (A) √50 [5√2 = √50]
8. Number of tangents drawn to a circle from an external point is:
(A) 1 (B) 2 (C) Infinite (D) 0
👉 Correct Answer: (B) 2
9. Value of tan 30° × tan 60° is:
(A) 0 (B) 1 (C) 2 (D) 1/2
👉 Correct Answer: (B) 1 [(1/√3) × √3 = 1]
10. If the perimeter and area of a circle are numerically equal, then radius r = ?
(A) 2 units (B) π units (C) 4 units (D) 7 units
👉 Correct Answer: (A) 2 units [2πr = πr2 ⇒ r = 2]
11. A card is drawn from a deck of 52. Probability of getting a red card is:
(A) 1/2 (B) 1/4 (C) 1/13 (D) 26/52
👉 Correct Answer: (A) 1/2 [26/52 = 1/2]
12. Mode of 2, 6, 4, 5, 0, 2, 1, 3, 2 is:
(A) 2 (B) 4 (C) 5 (D) 6
👉 Correct Answer: (A) 2
13. Volume of a sphere of radius r is:
(A) (4/3)πr3 (B) (2/3)πr3 (C) πr2h (D) 4πr2
👉 Correct Answer: (A) (4/3)πr3
14. Slope of line parallel to X-axis is:
(A) 0 (B) 1 (C) Undefined (D) -1
👉 Correct Answer: (A) 0
15. Product of roots of quadratic equation 2x2 - 3x + 1 = 0 is:
(A) 1/2 (B) 3/2 (C) -3/2 (D) -1/2
👉 Correct Answer: (A) 1/2 [c/a = 1/2]
16. The sum of first n natural numbers is given by:
(A) n(n+1)/2 (B) n(n-1)/2 (C) n2 (D) 2n
👉 Correct Answer: (A) n(n+1)/2
17. HCF of two prime numbers a and b is:
(A) 1 (B) ab (C) a (D) b
👉 Correct Answer: (A) 1
18. Total outcomes when two coins are tossed simultaneously:
(A) 2 (B) 4 (C) 6 (D) 8
👉 Correct Answer: (B) 4 [{HH, HT, TH, TT}]
19. In ΔABC, if DE || BC and AD=2 cm, DB=3 cm, AE=4 cm, then EC = ?
(A) 6 cm (B) 5 cm (C) 8 cm (D) 3 cm
👉 Correct Answer: (A) 6 cm [2/3 = 4/EC ⇒ EC = 6]
20. Probability of a sure event is:
(A) 0 (B) 1 (C) 1/2 (D) Infinite
👉 Correct Answer: (B) 1

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