TG SSC Class 10 Mathematics Official Model Paper 1

BOARD OF SECONDARY EDUCATION, TELANGANA

TG SSC CLASS 10 MATHEMATICS OFFICIAL MODEL PAPER 1

Full Question Paper & Step-by-Step Solutions | Muqeem Tutorial

TIME: 3 Hours 15 Mins PART - A & PART - B MAX MARKS: 80

PART - A (60 MARKS) — SUBJECTIVE QUESTIONS

SECTION - I (6 × 2 = 12 Marks) — Very Short Answer Questions (Answer ALL)

Q1. Express 420 as a product of its prime factors.

Solution:
• 420 = 2 × 210 = 2 × 2 × 105 = 2 × 2 × 3 × 35 = 2 × 2 × 3 × 5 × 7
👉 420 = 2² × 3 × 5 × 7

Q2. If A = {x : x is a prime number < 10} and B = {x : x is an odd number < 10}, find A ∩ B.

Solution:
• A = {2, 3, 5, 7}
• B = {1, 3, 5, 7, 9}
👉 A ∩ B = {3, 5, 7}

Q3. Find the sum and product of zeroes of the polynomial p(x) = 3x² - 5x + 2.

Solution:
• Here a = 3, b = -5, c = 2
• Sum of zeroes (α + β) = -b/a = -(-5)/3 = 5/3
• Product of zeroes (αβ) = c/a = 2/3
👉 Sum = 5/3, Product = 2/3

Q4. Find the 20th term of the AP: 4, 9, 14, 19...

Solution:
• First term a = 4, Common difference d = 9 - 4 = 5
• a₂₀ = a + 19d = 4 + 19(5) = 4 + 95 = 99
👉 a₂₀ = 99

Q5. Find the distance between the points (3, -2) and (-1, 1).

Solution:
• d = √[(-1 - 3)² + (1 - (-2))²] = √[(-4)² + 3²] = √[16 + 9] = √25 = 5
👉 Distance = 5 units

Q6. Evaluate: (sin 30° + tan 45°) / (cos 60° + cot 45°).

Solution:
• sin 30° = 1/2, tan 45° = 1, cos 60° = 1/2, cot 45° = 1
• Numerator = 1/2 + 1 = 3/2
• Denominator = 1/2 + 1 = 3/2
• Value = (3/2) / (3/2) = 1
👉 Value = 1

SECTION - II (6 × 4 = 24 Marks) — Short Answer Questions (Answer ALL)

Q7. Prove that 3 + √2 is an irrational number.

Solution:
1. Assume 3 + √2 = a/b (where a, b are co-prime integers, b ≠ 0).
2. √2 = (a/b) - 3 = (a - 3b) / b.
3. Since a and b are integers, (a - 3b)/b is rational, implying √2 is rational.
4. Contradiction! √2 is known to be irrational.
👉 Hence, 3 + √2 is Irrational. (Proved)

Q8. Find the zeroes of the quadratic polynomial 2x² - x - 6 and verify the relationship between zeroes and coefficients.

Solution:
• 2x² - 4x + 3x - 6 = 0 ⇒ 2x(x - 2) + 3(x - 2) = 0 ⇒ (2x + 3)(x - 2) = 0
• Zeroes: α = 2, β = -3/2.
• Sum α + β = 2 - 3/2 = 1/2 = -b/a = -(-1)/2 = 1/2. (Verified)
• Product αβ = 2 × (-3/2) = -3 = c/a = -6/2 = -3. (Verified)
👉 Zeroes: 2 and -3/2. Verified!

Q9. Solve the pair of linear equations by elimination method: 3x + 2y = 11 and 2x + 3y = 4.

Solution:
• Multiply 1st eq by 3: 9x + 6y = 33
• Multiply 2nd eq by 2: 4x + 6y = 8
• Subtracting: 5x = 25 ⇒ x = 5
• Substitute x = 5 in 3(5) + 2y = 11 ⇒ 15 + 2y = 11 ⇒ 2y = -4 ⇒ y = -2
👉 Solution: x = 5, y = -2

Q10. Find the ratio in which the Y-axis divides the line segment joining the points (5, -6) and (-1, -4).

Solution:
• Let the ratio be k : 1. Point on Y-axis is (0, y).
• x-coordinate: [k(-1) + 1(5)] / (k + 1) = 0 ⇒ -k + 5 = 0 ⇒ k = 5
👉 Required Ratio = 5 : 1

Q11. Two dice are thrown simultaneously. Find the probability that: (i) the sum is 7, (ii) the sum is a multiple of 5.

Solution:
• Total outcomes n(S) = 36
(i) Sum = 7: {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)} ⇒ 6 outcomes ⇒ P = 6/36 = 1/6
(ii) Sum is multiple of 5 (Sum = 5 or 10): {(1,4),(2,3),(3,2),(4,1),(4,6),(5,5),(6,4)} ⇒ 7 outcomes ⇒ P = 7/36
👉 Probabilities: (i) 1/6, (ii) 7/36

Q12. Prove that (1 + sec A) / sec A = sin² A / (1 - cos A).

Solution:
• LHS = (1 + 1/cos A) / (1/cos A) = 1 + cos A
• RHS = (1 - cos² A) / (1 - cos A) = [(1 - cos A)(1 + cos A)] / (1 - cos A) = 1 + cos A
👉 LHS = RHS = 1 + cos A. (Proved)

SECTION - III (4 × 6 = 24 Marks) — Essay Questions with Internal Choice

Q13 (A). Solve the system of linear equations graphically: x + y = 6 and x - y = 2.

Solution (13A):
• Line 1 (x + y = 6): Points (0, 6) and (6, 0)
• Line 2 (x - y = 2): Points (0, -2) and (2, 0)
• Graphing both lines, they intersect at point (4, 2).
👉 Solution: x = 4, y = 2

--- OR ---

Q13 (B). Find the zeroes of polynomial p(x) = x² - 5x + 6 graphically and verify.

Solution (13B):
• Factorizing: (x - 2)(x - 3) = 0. Parabola intersects X-axis at (2, 0) and (3, 0).
👉 Zeroes are x = 2 and x = 3

Q14 (A). State and prove Pythagoras Theorem.

Solution (14A):
• Statement: In a right-angled 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 toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 2 cm and the diameter of the base is 4 cm. Determine the volume of the toy. (Take π = 3.14)

Solution (14B):
• Radius r = 2 cm, Height of cone h = 2 cm.
• Vol(Toy) = Vol(Cone) + Vol(Hemisphere) = (1/3)πr²h + (2/3)πr³
• Vol(Toy) = (1/3)π(2²)(2) + (2/3)π(2³) = (8/3)π + (16/3)π = (24/3)π = 8π
• Vol(Toy) = 8 × 3.14 = 25.12 cm³
👉 Volume of Toy = 25.12 cm³

Q15 (A). From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

Solution (15A):
• Height of building = 7 m. Distance to tower d = 7 / tan 45° = 7 m.
• Height of tower above building h = d × tan 60° = 7√3 m.
• Total height of tower = 7 + 7√3 = 7(1 + √3) m.
👉 Height of Tower = 7(1 + √3) metres (≈ 19.12 m)

--- OR ---

Q15 (B). Draw a circle of radius 4 cm. From a point 10 cm away from its center, construct a pair of tangents to the circle and measure their lengths.

Solution (15B):
• OP = 10 cm, Radius r = 4 cm.
• Tangent Length = √(10² - 4²) = √(100 - 16) = √84 = 2√21 cm ≈ 9.16 cm.
1. Draw circle with center O and r = 4 cm. Mark point P at 10 cm.
2. Bisect OP to find midpoint M. Draw circle with center M and radius MO.
3. Let it intersect original circle at A and B. Join PA and PB.
👉 PA = PB ≈ 9.16 cm.

Q16 (A). Calculate the Mean for the following frequency distribution:

Class Interval	0 - 10	10 - 20	20 - 30	30 - 40	40 - 50
Frequency (fi)	4	8	12	10	6

Solution (16A):
• Mid-values (x_i) = 5, 15, 25, 35, 45
• f_ix_i = 20, 120, 300, 350, 270 ⇒ Σf_ix_i = 1060, Σf_i = 40
• Mean = Σf_ix_i / Σf_i = 1060 / 40 = 26.5
👉 Mean = 26.5

--- OR ---

Q16 (B). One card is drawn from a well-shuffled deck of 52 cards. Find probability of getting: (i) a face card, (ii) a red king, (iii) a spade, (iv) jack of hearts, (v) a non-face card, (vi) a red queen.

Solution (16B):
(i) Face cards = 12 ⇒ P = 12/52 = 3/13
(ii) Red Kings = 2 ⇒ P = 2/52 = 1/26
(iii) Spades = 13 ⇒ P = 13/52 = 1/4
(iv) Jack of hearts = 1 ⇒ P = 1/52
(v) Non-face cards = 40 ⇒ P = 40/52 = 10/13
(vi) Red Queens = 2 ⇒ P = 2/52 = 1/26
👉 Probabilities: (i) 3/13, (ii) 1/26, (iii) 1/4, (iv) 1/52, (v) 10/13, (vi) 1/26

PART - B (20 MARKS) — MULTIPLE CHOICE QUESTIONS (1 Mark Each)

1. The HCF of 96 and 404 is:

(A) 2      (B) 4      (C) 8      (D) 12

👉 Correct Answer: (B) 4

2. If A = {1, 2, 3} and B = {3, 4, 5}, then n(A ∪ B) is:

(A) 5      (B) 6      (C) 4      (D) 3

👉 Correct Answer: (A) 5 [A ∪ B = {1, 2, 3, 4, 5}]

3. Number of zeroes of polynomial p(x) = x³ - x is:

(A) 1      (B) 2      (C) 3      (D) 0

👉 Correct Answer: (C) 3 [x(x-1)(x+1) = 0]

4. The value of k for which x + 2y = 3 and 5x + ky + 7 = 0 has no solution is:

(A) 10      (B) 5      (C) 2      (D) -10

👉 Correct Answer: (A) 10 [1/5 = 2/k ⇒ k = 10]

5. Nature of roots of 2x² - 3x + 5 = 0:

(A) Real & Equal      (B) Real & Distinct      (C) No Real Roots      (D) Rational

👉 Correct Answer: (C) No Real Roots [D = 9 - 40 = -31 < 0]

6. 10th term of AP 5, 8, 11, 14... is:

(A) 32      (B) 35      (C) 38      (D) 30

👉 Correct Answer: (A) 32 [5 + 9(3) = 32]

7. Distance between points (a, 0) and (0, a) is:

(A) a      (B) a√2      (C) 2a      (D) a²

👉 Correct Answer: (B) a√2 [√(a² + a²) = a√2]

8. Midpoint of line segment joining (-2, 4) and (6, 8) is:

(A) (2, 6)      (B) (4, 6)      (C) (2, 4)      (D) (4, 4)

👉 Correct Answer: (A) (2, 6) [(-2+6)/2=2, (4+8)/2=6]

9. Value of sin² 45° + cos² 45° is:

(A) 0      (B) 1      (C) 1/2      (D) 1/√2

👉 Correct Answer: (B) 1

10. If tan θ = 1, then θ = ?

(A) 30°      (B) 45°      (C) 60°      (D) 90°

👉 Correct Answer: (B) 45°

11. Length of shadow of a pole 10 m high when sun elevation is 45°:

(A) 5 m      (B) 10 m      (C) 10√3 m      (D) 10/√3 m

👉 Correct Answer: (B) 10 m

12. Number of tangents to a circle from a point on the circle is:

(A) 0      (B) 1      (C) 2      (D) Infinite

👉 Correct Answer: (B) 1

13. Curved surface area of a cylinder of radius r and height h is:

(A) 2πrh      (B) πr²h      (C) 2πr(r+h)      (D) πrh

👉 Correct Answer: (A) 2πrh

14. Probability of getting an odd number on rolling a fair die once is:

(A) 1/2      (B) 1/3      (C) 1/6      (D) 2/3

👉 Correct Answer: (A) 1/2 [3/6 = 1/2]

15. If P(E) = 0.08, then P(NOT E) = ?

(A) 0.92      (B) 0.08      (C) 0.82      (D) 1.08

👉 Correct Answer: (A) 0.92 [1 - 0.08 = 0.92]

16. Mode of data: 4, 6, 7, 6, 8, 6, 9 is:

(A) 6      (B) 7      (C) 8      (D) 9

👉 Correct Answer: (A) 6

17. Class mark of class interval 30 - 50 is:

(A) 30      (B) 40      (C) 50      (D) 80

👉 Correct Answer: (B) 40 [(30+50)/2 = 40]

18. In ΔABC, if ∠B = 90°, AB = 6 cm, BC = 8 cm, then AC = ?

(A) 10 cm      (B) 12 cm      (C) 14 cm      (D) 100 cm

👉 Correct Answer: (A) 10 cm [√(36 + 64) = 10]

19. Slope of line 2x + 3y = 6 is:

(A) 2/3      (B) -2/3      (C) 3/2      (D) -3/2

👉 Correct Answer: (B) -2/3 [-a/b = -2/3]

20. Sum of first 10 terms of AP 2, 4, 6, 8... is:

(A) 100      (B) 110      (C) 120      (D) 90

👉 Correct Answer: (B) 110 [(10/2)(4 + 18) = 110]