BOARD OF SECONDARY EDUCATION, TELANGANA
TS SSC CLASS 10 MATHEMATICS PUBLIC EXAMINATION 2021
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 lemma to find the HCF of 90 and 144.
Solution:
By Euclid's Division Lemma: a = bq + r
1. 144 = 90 × 1 + 54
2. 90 = 54 × 1 + 36
3. 54 = 36 × 1 + 18
4. 36 = 18 × 2 + 0
Since remainder is 0, the divisor at this final step is 18.
👉 HCF(90, 144) = 18
By Euclid's Division Lemma: a = bq + r
1. 144 = 90 × 1 + 54
2. 90 = 54 × 1 + 36
3. 54 = 36 × 1 + 18
4. 36 = 18 × 2 + 0
Since remainder is 0, the divisor at this final step is 18.
👉 HCF(90, 144) = 18
Q2. If A = {1, 2, 3, 4, 5} and B = {2, 4, 6, 8}, find A - B and B - A.
Solution:
• A - B = {1, 3, 5}
• B - A = {6, 8}
👉 A - B = {1, 3, 5}, B - A = {6, 8}
• A - B = {1, 3, 5}
• B - A = {6, 8}
👉 A - B = {1, 3, 5}, B - A = {6, 8}
Q3. Find the zeroes of the polynomial p(x) = x2 - 7x + 12.
Solution:
• x2 - 4x - 3x + 12 = 0 ⇒ (x - 4)(x - 3) = 0
👉 Zeroes are x = 4 and x = 3
• x2 - 4x - 3x + 12 = 0 ⇒ (x - 4)(x - 3) = 0
👉 Zeroes are x = 4 and x = 3
Q4. Find the 15th term of the AP: 3, 8, 13, 18...
Solution:
• a = 3, d = 5
• a15 = 3 + 14(5) = 3 + 70 = 73
👉 a15 = 73
• a = 3, d = 5
• a15 = 3 + 14(5) = 3 + 70 = 73
👉 a15 = 73
Q5. Find the distance between the points A(3, 4) and B(7, 7).
Solution:
• d = √[(7 - 3)2 + (7 - 4)2] = √[16 + 9] = √25 = 5
👉 Distance = 5 units
• d = √[(7 - 3)2 + (7 - 4)2] = √[16 + 9] = √25 = 5
👉 Distance = 5 units
Q6. If tan θ = 3/4, find sin θ and cos θ.
Solution:
• Opposite = 3, Adjacent = 4, Hypotenuse = 5
👉 sin θ = 3/5, cos θ = 4/5
• Opposite = 3, Adjacent = 4, Hypotenuse = 5
👉 sin θ = 3/5, cos θ = 4/5
SECTION - II (6 × 4 = 24 Marks) — Short Answer Questions (Answer ALL)
Q7. Prove that 2 + √3 is an irrational number.
Solution:
1. Let 2 + √3 = a/b (a, b coprime integers, b ≠ 0).
2. √3 = (a - 2b) / b.
3. Since a, b are integers, (a - 2b)/b is rational, implying √3 is rational.
4. Contradiction! √3 is irrational.
👉 Hence, 2 + √3 is Irrational. (Proved)
1. Let 2 + √3 = a/b (a, b coprime integers, b ≠ 0).
2. √3 = (a - 2b) / b.
3. Since a, b are integers, (a - 2b)/b is rational, implying √3 is rational.
4. Contradiction! √3 is irrational.
👉 Hence, 2 + √3 is Irrational. (Proved)
Q8. Find the zeroes of quadratic polynomial x2 + 7x + 10 and verify relationship between zeroes and coefficients.
Solution:
• (x + 5)(x + 2) = 0 ⇒ α = -5, β = -2.
• Sum α + β = -7 = -b/a = -7/1. (Verified)
• Product αβ = 10 = c/a = 10/1. (Verified)
👉 Zeroes: -5 and -2. Verified!
• (x + 5)(x + 2) = 0 ⇒ α = -5, β = -2.
• Sum α + β = -7 = -b/a = -7/1. (Verified)
• Product αβ = 10 = c/a = 10/1. (Verified)
👉 Zeroes: -5 and -2. Verified!
Q9. Solve by elimination method: 2x + 3y = 11 and 2x - 4y = -24.
Solution:
• Subtracting equations: 7y = 35 ⇒ y = 5
• Substituting y = 5 in 2x + 3(5) = 11 ⇒ 2x = -4 ⇒ x = -2
👉 Solution: x = -2, y = 5
• Subtracting equations: 7y = 35 ⇒ y = 5
• Substituting y = 5 in 2x + 3(5) = 11 ⇒ 2x = -4 ⇒ x = -2
👉 Solution: x = -2, y = 5
Q10. Find the coordinates of the point which divides line segment joining (4, -3) and (8, 5) in ratio 3 : 1 internally.
Solution:
• P(x, y) = [(3(8) + 1(4))/4, (3(5) + 1(-3))/4] = [28/4, 12/4] = (7, 3)
👉 Required Point P = (7, 3)
• P(x, y) = [(3(8) + 1(4))/4, (3(5) + 1(-3))/4] = [28/4, 12/4] = (7, 3)
👉 Required Point P = (7, 3)
Q11. A card is drawn from a well-shuffled deck of 52 cards. Find probability of getting: (i) a king of red color, (ii) a face card.
Solution:
(i) Red Kings = 2 ⇒ P = 2/52 = 1/26
(ii) Face cards = 12 ⇒ P = 12/52 = 3/13
👉 Probabilities: (i) 1/26, (ii) 3/13
(i) Red Kings = 2 ⇒ P = 2/52 = 1/26
(ii) Face cards = 12 ⇒ P = 12/52 = 3/13
👉 Probabilities: (i) 1/26, (ii) 3/13
Q12. Prove that (1 + tan2 A) / (1 + cot2 A) = tan2 A.
Solution:
LHS = sec2 A / csc2 A = (1/cos2 A) / (1/sin2 A) = sin2 A / cos2 A = tan2 A = RHS
👉 Hence Proved.
LHS = sec2 A / csc2 A = (1/cos2 A) / (1/sin2 A) = sin2 A / cos2 A = tan2 A = RHS
👉 Hence Proved.
SECTION - III (4 × 6 = 24 Marks) — Essay Questions with Internal Choice
Q13 (A). Solve linear equations graphically: 2x + y = 6 and 2x - y = 2.
Solution (13A):
• Line 1: (0, 6) and (3, 0)
• Line 2: (0, -2) and (1, 0)
• Intersection point on graph = (2, 2).
👉 Solution: x = 2, y = 2
• Line 1: (0, 6) and (3, 0)
• Line 2: (0, -2) and (1, 0)
• Intersection point on graph = (2, 2).
👉 Solution: x = 2, y = 2
--- OR ---
Q13 (B). Draw the graph of polynomial p(x) = x2 - 3x - 4 and find its zeroes.
Solution (13B):
• Parabola intersects X-axis at (-1, 0) and (4, 0).
👉 Zeroes are x = -1 and x = 4
• Parabola intersects X-axis at (-1, 0) and (4, 0).
👉 Zeroes are x = -1 and x = 4
Q14 (A). State and prove Pythagoras Theorem.
Solution (14A):
• Statement: In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
• Proof: In ΔABC right-angled at B, draw BD ⊥ AC.
ΔABD ~ ΔABC ⇒ AB2 = AC × AD --- (1)
ΔBCD ~ ΔABC ⇒ BC2 = AC × CD --- (2)
Adding (1) and (2): AB2 + BC2 = AC(AD + CD) = AC2.
👉 AB2 + BC2 = AC2. (Proved)
• Statement: In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
• Proof: In ΔABC right-angled at B, draw BD ⊥ AC.
ΔABD ~ ΔABC ⇒ AB2 = AC × AD --- (1)
ΔBCD ~ ΔABC ⇒ BC2 = AC × CD --- (2)
Adding (1) and (2): AB2 + BC2 = AC(AD + CD) = AC2.
👉 AB2 + BC2 = AC2. (Proved)
--- OR ---
Q14 (B). A solid metallic sphere of radius 10.5 cm is melted and recast into smaller solid cones, each of radius 3.5 cm and height 3 cm. Find the total number of cones formed.
Solution (14B):
• Volume of Sphere = (4/3)π(10.5)3 = 1543.5π cm3
• Volume of 1 Cone = (1/3)π(3.5)2(3) = 12.25π cm3
• Cones formed = 1543.5π / 12.25π = 126
👉 Total Cones Formed = 126
• Volume of Sphere = (4/3)π(10.5)3 = 1543.5π cm3
• Volume of 1 Cone = (1/3)π(3.5)2(3) = 12.25π cm3
• Cones formed = 1543.5π / 12.25π = 126
👉 Total Cones Formed = 126
Q15 (A). A tree breaks due to a storm and the broken part bends so that the top touches the ground making an angle of 30° with it. The distance from foot to top is 8 m. Find tree height.
Solution (15A):
• tan 30° = h / 8 ⇒ h = 8 / √3 m.
• cos 30° = 8 / x ⇒ x = 16 / √3 m.
• Total height = h + x = 24 / √3 = 8√3 m.
👉 Tree Height = 8√3 metres (≈ 13.85 m)
• tan 30° = h / 8 ⇒ h = 8 / √3 m.
• cos 30° = 8 / x ⇒ x = 16 / √3 m.
• Total height = h + x = 24 / √3 = 8√3 m.
👉 Tree Height = 8√3 metres (≈ 13.85 m)
--- OR ---
Q15 (B). Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Solution (15B):
• Radius ⊥ Tangent ⇒ ∠OAL1 = 90° and ∠OBL2 = 90°.
• Since alternate interior angles are equal (90° = 90°),
👉 Tangents L1 and L2 are parallel. (Proved)
• Radius ⊥ Tangent ⇒ ∠OAL1 = 90° and ∠OBL2 = 90°.
• Since alternate interior angles are equal (90° = 90°),
👉 Tangents L1 and L2 are parallel. (Proved)
Q16 (A). Calculate the Mean and Mode for the following frequency distribution:
| Class Interval | 0 - 10 | 10 - 20 | 20 - 30 | 30 - 40 | 40 - 50 |
|---|---|---|---|---|---|
| Frequency (fi) | 5 | 8 | 15 | 10 | 2 |
Solution (16A):
• Mean = Σfixi / Σfi = 960 / 40 = 24
• Mode = 20 + [(15 - 8)/(30 - 8 - 10)] × 10 = 20 + 5.83 = 25.83
👉 Mean = 24, Mode = 25.83
• Mean = Σfixi / Σfi = 960 / 40 = 24
• Mode = 20 + [(15 - 8)/(30 - 8 - 10)] × 10 = 20 + 5.83 = 25.83
👉 Mean = 24, Mode = 25.83
--- OR ---
Q16 (B). Two dice are thrown at the same time. Find the probability of getting: (i) a doublet, (ii) sum = 8, (iii) sum > 10.
Solution (16B):
(i) Doublet = 6/36 = 1/6
(ii) Sum = 8 = 5/36
(iii) Sum > 10 = 3/36 = 1/12
👉 Probabilities: (i) 1/6, (ii) 5/36, (iii) 1/12
(i) Doublet = 6/36 = 1/6
(ii) Sum = 8 = 5/36
(iii) Sum > 10 = 3/36 = 1/12
👉 Probabilities: (i) 1/6, (ii) 5/36, (iii) 1/12
PART - B (20 MARKS) — MULTIPLE CHOICE QUESTIONS (1 Mark Each)
1. The value of log2 32 is equal to:
(A) 2 (B) 5 (C) 10 (D) 32
👉 Correct Answer: (B) 5 [25 = 32]
2. If log10 (x - 2) = 1, then x = ?
(A) 8 (B) 10 (C) 12 (D) 20
👉 Correct Answer: (C) 12 [x - 2 = 101 ⇒ x = 12]
3. The slope of a line parallel to the X-axis is:
(A) 0 (B) 1 (C) Undefined (D) -1
👉 Correct Answer: (A) 0
4. Value of tan 30° × tan 60° is:
(A) 0 (B) 1 (C) 2 (D) 1/2
👉 Correct Answer: (B) 1 [(1/√3) × √3 = 1]
5. The mode of 2, 3, 2, 4, 5, 2, 3 is:
(A) 3 (B) 2 (C) 4 (D) 5
👉 Correct Answer: (B) 2
6. Probability of an impossible event is:
(A) 0 (B) 1 (C) 1/2 (D) -1
👉 Correct Answer: (A) 0
7. If P(E) = 0.05, then P(NOT E) is:
(A) 0.95 (B) 0.50 (C) 0.05 (D) 1.05
👉 Correct Answer: (A) 0.95
8. Total surface area of a hemisphere of radius r is:
(A) 2πr2 (B) 3πr2 (C) 4πr2 (D) πr2
👉 Correct Answer: (B) 3πr2
9. Midpoint of class interval 10 - 20 is:
(A) 10 (B) 15 (C) 20 (D) 30
👉 Correct Answer: (B) 15
10. Distance of point (3, 4) from origin is:
(A) 3 (B) 4 (C) 5 (D) 7
👉 Correct Answer: (C) 5
11. HCF of 8, 9, 25 is:
(A) 1 (B) 2 (C) 4 (D) 25
👉 Correct Answer: (A) 1
12. Product of zeroes of 2x2 - 8x + 6 is:
(A) 4 (B) -4 (C) 3 (D) -3
👉 Correct Answer: (C) 3
13. Common difference of AP: 5, 2, -1, -4... is:
(A) 3 (B) -3 (C) 2 (D) -2
👉 Correct Answer: (B) -3
14. Value of tan 45° + cot 45° is:
(A) 1 (B) 2 (C) 0 (D) √2
👉 Correct Answer: (B) 2
15. Discriminant of x2 - 4x + 4 = 0 is:
(A) 0 (B) 16 (C) 4 (D) -4
👉 Correct Answer: (A) 0
16. Angle of elevation of sun when height equals shadow length:
(A) 30° (B) 45° (C) 60° (D) 90°
👉 Correct Answer: (B) 45°
17. Volume of a cone formula:
(A) πr2h (B) (1/3)πr2h (C) (4/3)πr3 (D) 2πrh
👉 Correct Answer: (B) (1/3)πr2h
18. Probability of getting Head in a single coin toss:
(A) 0 (B) 1 (C) 1/2 (D) 1/4
👉 Correct Answer: (C) 1/2
19. Empirical relation: Mode = 3 Median - 2 Mean
(A) Mode = 3 Median - 2 Mean (B) Mode = 2 Median - 3 Mean (C) Mode = Median + Mean (D) None
👉 Correct Answer: (A) Mode = 3 Median - 2 Mean
20. Which of the following is NOT a measure of central tendency?
(A) Mean (B) Median (C) Range (D) Mode
👉 Correct Answer: (C) Range

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