Class 10 Board Examination — Mathematics Short Answer Questions (2 Marks Each)
Q1. Find the HCF of 96 and 404 using the prime factorisation method.
Q2. Find the zeroes of the polynomial p(x) = x² − 3x − 10.
Q3. Solve the following pair of linear equations by the substitution method: x + y = 14 x − y = 4
Q4. Find the roots of the quadratic equation 2x² + x − 6 = 0 using the factorisation method.
Q5. How many terms of the AP: 9, 17, 25, ... must be taken so that their sum is 636?
Q6. Find the coordinates of the point which divides the line segment joining the points A(1, 3) and B(4, 6) in the ratio 2 : 1 internally.
Q7. In △ABC, DE ∥ BC, where D is on AB and E is on AC. If AD = 3 cm, DB = 5 cm and AE = 4.5 cm, find EC.
Q8. Prove that: (sin θ + cos θ)² + (sin θ − cos θ)² = 2
Q9. Find the area of a sector of a circle with radius 6 cm and angle 60°. (Use π = 3.14)
Q10. Two coins are tossed simultaneously. Find the probability of getting: (i) exactly one head (ii) at least one head
Q11. Find the 20th term of the AP: 2, 7, 12, 17, ...
Q12. Show that 5 + √2 is irrational.
Q13. Find the value of k for which the quadratic equation kx² − 6x + 1 = 0 has real and equal roots.
Q14. Find the distance of the point P(3, 4) from the origin.
Q15. A bag contains 4 red, 5 blue, and 3 green balls. A ball is drawn at random. Find the probability that the ball drawn is: (i) not red (ii) green
ANSWERS
A1. 96 = 2⁵ × 3 404 = 2² × 101 HCF = 2² = 4
A2. x² − 3x − 10 = 0 (x − 5)(x + 2) = 0 x = 5 and x = −2
A3. x + y = 14 → x = 14 − y Substituting: (14 − y) − y = 4 14 − 2y = 4 → y = 5 x = 14 − 5 = 9 x = 9, y = 5
A4. 2x² + x − 6 = 0 2x² + 4x − 3x − 6 = 0 2x(x + 2) − 3(x + 2) = 0 (2x − 3)(x + 2) = 0 x = 3/2 and x = −2
A5. a = 9, d = 8, Sₙ = 636 Sₙ = n/2 [2a + (n−1)d] 636 = n/2 [18 + 8(n−1)] 1272 = n(8n + 10) 8n² + 10n − 1272 = 0 4n² + 5n − 636 = 0 (n − 12)(4n + 53) = 0 n = 12
A6. Using section formula, m:n = 2:1, A(1,3), B(4,6) x = (2×4 + 1×1)/(2+1) = 9/3 = 3 y = (2×6 + 1×3)/(2+1) = 15/3 = 5 Point = (3, 5)
A7. By BPT: AD/DB = AE/EC 3/5 = 4.5/EC EC = 7.5 cm
A8. LHS = (sin θ + cos θ)² + (sin θ − cos θ)² = sin²θ + 2sinθcosθ + cos²θ + sin²θ − 2sinθcosθ + cos²θ = 1 + 1 = 2 = RHS ✓
A9. Area of sector = (θ/360°) × πr² = (60/360) × 3.14 × 6 × 6 = (1/6) × 113.04 = 18.84 cm²
A10. Sample space = {HH, HT, TH, TT}, n(S) = 4 (i) Exactly one head: {HT, TH} → P = 2/4 = 1/2 (ii) At least one head: {HH, HT, TH} → P = 3/4
A11. a = 2, d = 5, n = 20 a₂₀ = a + (n−1)d = 2 + 19 × 5 = 2 + 95 = 97
A12. Assume 5 + √2 is rational. Then √2 = (5 + √2) − 5 = rational − rational = rational. But √2 is irrational — a contradiction. Hence, 5 + √2 is irrational. ✓
A13. For real and equal roots: b² − 4ac = 0 (−6)² − 4(k)(1) = 0 36 − 4k = 0 k = 9
A14. Distance from origin = √(x² + y²) = √(3² + 4²) = √(9 + 16) = √25 = 5 units
A15. Total balls = 4 + 5 + 3 = 12 (i) P(not red) = (5+3)/12 = 8/12 = 2/3 (ii) P(green) = 3/12 = 1/4
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