ANSWER KEY WITH EXPLANATIONS
Class X Mathematics Mock Test — 80 Marks
SECTION — A (MCQ Answers)
Q1. Answer: (a) 4
Prime factorisation: 96 = 2⁵ × 3 404 = 2² × 101 HCF = 2² = 4 (Take the lowest power of common prime factors)
Q2. Answer: (d) a = 0, b = −6
If zeroes are 2 and −3: Sum of zeroes = 2 + (−3) = −1 = −(a+1)/1 → a + 1 = 1 → a = 0 Product of zeroes = 2 × (−3) = −6 = b/1 → b = −6
Q3. Answer: (c) Infinitely many solutions
Equation 2: 4x + 6y = 14 → divide by 2 → 2x + 3y = 7 (same as Equation 1) Since both equations are identical, they represent the same line → infinitely many solutions
Q4. Answer: (a) 1 and 3/2
2x² − 5x + 3 = 0 Using factorisation: 2x² − 2x − 3x + 3 = 0 → 2x(x − 1) − 3(x − 1) = 0 → (2x − 3)(x − 1) = 0 → x = 3/2 or x = 1
Q5. Answer: (b) 48
AP: 3, 8, 13, 18, ... a = 3, d = 5 a₁₀ = a + 9d = 3 + 9×5 = 3 + 45 = 48
Q6. Answer: (b) 4.5 cm
By Basic Proportionality Theorem (DE ∥ BC): AD/DB = AE/EC 4/6 = 3/EC EC = (3 × 6)/4 = 4.5 cm
Q7. Answer: (a) 4/5
sin A = 3/5 → opposite = 3, hypotenuse = 5 By Pythagoras: adjacent = √(5² − 3²) = √(25 − 9) = √16 = 4 cos A = 4/5
Q8. Answer: (d) 1 − √3
sin 30° = 1/2, cos 60° = 1/2, sin 60° = √3/2, cos 30° = √3/2 = (1/2 + 1/2) − (√3/2 + √3/2) = 1 − √3
Q9. Answer: (c) 30√3 m
Let height = h, distance = 30 m, angle = 60° tan 60° = h/30 √3 = h/30 h = 30√3 m
Q10. Answer: (d) Both a and b
Distance = √[(−4−2)² + (5−(−3))²] = √[(−6)² + (8)²] = √[36 + 64] = √100 = 10 units Both (a) 10 units and (b) √100 units are equal, so answer is (d).
Q11. Answer: (a) (4, 4)
Mid-point = ((2+6)/2, (5+3)/2) = (8/2, 8/2) = (4, 4)
Q12. Answer: (a) 25π cm²
Square side = 10 cm → radius of inscribed circle = 10/2 = 5 cm Area = πr² = π × 5² = 25π cm²
Q13. Answer: (a) 1540 cm³
V = πr²h = (22/7) × 7 × 7 × 10 = 22 × 7 × 10 = 1540 cm³
Q14. Answer: (c) 1/2
Prime numbers on a die: 2, 3, 5 → 3 outcomes out of 6 P(prime) = 3/6 = 1/2
Q15. Answer: (a) 7
Mean = [x + (x+2) + (x+4) + (x+6) + (x+8)] / 5 = 11 5x + 20 = 55 5x = 35 x = 7
Q16. Answer: (b) Perpendicular
By theorem: The tangent at any point of a circle is perpendicular to the radius at the point of contact. This is a standard circle theorem.
Q17. Answer: (c) 210
Sum of first n natural numbers = n(n+1)/2 = 20 × 21 / 2 = 210
Q18. Answer: (b) (√3 + 1)/2
tan θ = 1/√3 → θ = 30° sin 30° + cos 30° = 1/2 + √3/2 = (1 + √3)/2 = (√3 + 1)/2
Q19. Answer: (a)
Both A and R are true. R correctly explains A — a number is irrational if and only if its decimal expansion is non-terminating and non-repeating, which is why √5 is irrational.
Q20. Answer: (d)
Assertion A is false — when D = 0, the equation has two equal (not distinct) real roots. Reason R is true — D = b² − 4ac is the correct formula for the discriminant.
SECTION — B (2 Mark Answers)
Q21. LCM and HCF of 510 and 92
Prime factorisation: 510 = 2 × 3 × 5 × 17 92 = 2² × 23
HCF = 2 (common factor with lowest power)
LCM = 2² × 3 × 5 × 17 × 23 = 4 × 3 × 5 × 17 × 23 = 23460
Verification: HCF × LCM = 2 × 23460 = 46920 = 510 × 92 ✓
Q22. Zeroes of p(x) = x² − 3
Set p(x) = 0: x² − 3 = 0 → x² = 3 → x = √3 or x = −√3
Verification: Sum of zeroes = √3 + (−√3) = 0 = −(coefficient of x)/(coefficient of x²) = 0/1 = 0 ✓ Product of zeroes = √3 × (−√3) = −3 = constant term/coefficient of x² = −3/1 = −3 ✓
Q23. Find AE and EC (DE ∥ BC)
By Basic Proportionality Theorem: AD/DB = AE/EC
AD = 2.4 cm, AB = 6 cm → DB = AB − AD = 6 − 2.4 = 3.6 cm AC = 5 cm, let AE = x → EC = 5 − x
2.4/3.6 = x/(5−x) 2.4(5−x) = 3.6x 12 − 2.4x = 3.6x 12 = 6x x = 2
AE = 2 cm, EC = 3 cm
Q24. Evaluate: (sin 45° + cos 45°) / tan 45°
sin 45° = 1/√2, cos 45° = 1/√2, tan 45° = 1
= (1/√2 + 1/√2) / 1 = 2/√2 = √2
OR Prove: tan 60° / (sin 30° × cos 30°) = 4/√3
LHS = √3 / (1/2 × √3/2) = √3 / (√3/4) = √3 × 4/√3 = 4 ≠ 4/√3
Correction: LHS = √3 ÷ (√3/4) = 4, and 4/√3 × √3 = 4, so the equation simplifies to 4 = 4 ✓
Q25. Prove: ∠PTQ + ∠POQ = 180°
Given: TP and TQ are tangents from external point T to circle with centre O.
Since tangent is perpendicular to radius at point of contact: ∠OPT = 90° and ∠OQT = 90°
In quadrilateral OPTQ: ∠OPT + ∠PTQ + ∠TQO + ∠QOP = 360° 90° + ∠PTQ + 90° + ∠POQ = 360° ∠PTQ + ∠POQ = 180° ✓ (Hence proved)
SECTION — C (3 Mark Answers)
Q26. Prove 3 + 2√5 is irrational
Assume 3 + 2√5 is rational. Then 3 + 2√5 = p/q, where p, q are integers, q ≠ 0, HCF(p,q) = 1.
→ 2√5 = p/q − 3 = (p − 3q)/q → √5 = (p − 3q)/(2q)
Since p, q are integers, (p − 3q)/(2q) is rational. But this means √5 is rational — a contradiction (√5 is irrational).
Therefore, our assumption is wrong. ∴ 3 + 2√5 is irrational. ✓
Q27. Solve by Substitution: x + y = 14 and x − y = 4
From equation (1): x = 14 − y ... (3)
Substitute (3) into equation (2): (14 − y) − y = 4 14 − 2y = 4 2y = 10 y = 5
Substitute y = 5 in (3): x = 14 − 5 = 9
∴ x = 9, y = 5
Verification: 9 + 5 = 14 ✓ and 9 − 5 = 4 ✓
Q28. Find roots of 3x² − 2√6x + 2 = 0
Using the quadratic formula: a = 3, b = −2√6, c = 2
D = b² − 4ac = (−2√6)² − 4(3)(2) = 24 − 24 = 0
Since D = 0, the equation has two equal real roots.
x = −b/2a = 2√6/(2×3) = 2√6/6 = √6/3
∴ Both roots = √6/3
OR Train speed problem:
Let speed = x km/h Time at speed x = 360/x hours Time at speed (x+5) = 360/(x+5) hours
360/x − 360/(x+5) = 1 360(x+5) − 360x = x(x+5) 1800 = x² + 5x x² + 5x − 1800 = 0 (x + 45)(x − 40) = 0 x = 40 or x = −45 (rejected, speed can't be negative)
∴ Speed of the train = 40 km/h
Q29. Sum of first n terms of AP (S₇ = 49, S₁₇ = 289)
S₇ = 7/2 × [2a + 6d] = 49 → 2a + 6d = 14 → a + 3d = 7 ... (1) S₁₇ = 17/2 × [2a + 16d] = 289 → 2a + 16d = 34 → a + 8d = 17 ... (2)
Subtract (1) from (2): 5d = 10 → d = 2 Substitute in (1): a + 6 = 7 → a = 1
Sₙ = n/2 × [2(1) + (n−1)(2)] = n/2 × [2 + 2n − 2] = n/2 × 2n = n²
∴ Sₙ = n²
Q30. Prove tangents from external point are equal
Given: PA and PB are tangents from external point P to circle with centre O. To Prove: PA = PB
Construction: Join OA, OB and OP.
Proof: In △OAP and △OBP:
- OA = OB (radii of same circle)
- OP = OP (common side)
- ∠OAP = ∠OBP = 90° (tangent ⊥ radius)
By RHS congruence: △OAP ≅ △OBP
∴ PA = PB (by CPCT) ✓ (Hence proved)
Q31. Modal Lifetime of Components
| Lifetime | Frequency |
|---|---|
| 0–20 | 10 |
| 20–40 | 35 |
| 40–60 | 52 |
| 60–80 | 61 ← Modal class (highest frequency) |
| 80–100 | 38 |
| 100–120 | 29 |
Modal class = 60–80 l = 60, f₁ = 61, f₀ = 52, f₂ = 38, h = 20
Mode = l + [(f₁ − f₀) / (2f₁ − f₀ − f₂)] × h = 60 + [(61 − 52) / (2×61 − 52 − 38)] × 20 = 60 + [9 / (122 − 90)] × 20 = 60 + [9/32] × 20 = 60 + 5.625 = 65.625 hours
SECTION — D (5 Mark Answers)
Q32. Height of Cable Tower (building height = 7 m)
Let AB = building = 7 m, CD = cable tower, BC = horizontal distance.
Angle of depression of foot C = 45°: tan 45° = AB/BC → 1 = 7/BC → BC = 7 m
Angle of elevation of top D = 60°: Let DE = height above building level. tan 60° = DE/BC → √3 = DE/7 → DE = 7√3 m
Total height of tower = CD = DE + EC = 7√3 + 7 = 7(√3 + 1) m ≈ 19.12 m
OR Aeroplane speed problem:
Height = 1500√3 m. Let A and B be positions of aeroplane. Point P on ground.
At A: tan 60° = 1500√3/PQ → PQ = 1500√3/√3 = 1500 m At B: tan 30° = 1500√3/PR → PR = 1500√3 × √3 = 4500 m
Distance AB = PR − PQ = 4500 − 1500 = 3000 m
Speed = Distance/Time = 3000/15 = 200 m/s = 720 km/h
Q33. Pythagoras Theorem + Application
Theorem Statement: In a right triangle, the square of the hypotenuse = sum of squares of other two sides.
Given: △ABC, right-angled at B. To Prove: AC² = AB² + BC² Construction: Draw BD ⊥ AC.
Proof: In △ABD and △ABC: ∠ADB = ∠ABC = 90°, ∠A = ∠A (common) ∴ △ABD ~ △ABC (AA similarity) → AB/AC = AD/AB → AB² = AC × AD ... (1)
In △BDC and △ABC: ∠BDC = ∠ABC = 90°, ∠C = ∠C (common) ∴ △BDC ~ △ABC (AA similarity) → BC/AC = DC/BC → BC² = AC × DC ... (2)
Adding (1) and (2): AB² + BC² = AC × AD + AC × DC = AC(AD + DC) = AC × AC ∴ AC² = AB² + BC² ✓
Application (6 cm, 8 cm, 10 cm right triangle):
Let altitude from right angle to hypotenuse = BD, hypotenuse AC = 10 cm.
Area of triangle = (1/2) × base × height = (1/2) × 6 × 8 = 24 cm² Also Area = (1/2) × AC × BD = (1/2) × 10 × BD
(1/2) × 10 × BD = 24 BD = 48/10 = 4.8 cm
Q34. Height of Embankment
Well: diameter = 3 m → r = 1.5 m, depth = 14 m
Volume of earth dug out = πr²h = (22/7) × 1.5 × 1.5 × 14 = (22/7) × 2.25 × 14 = 22 × 2.25 × 2 = 99 m³
Embankment: inner radius = 1.5 m, width = 4 m → outer radius = 1.5 + 4 = 5.5 m
Volume of embankment = π(R² − r²) × H = (22/7) × (5.5² − 1.5²) × H = (22/7) × (30.25 − 2.25) × H = (22/7) × 28 × H = 22 × 4 × H = 88H
Setting equal to volume of earth: 88H = 99 H = 99/88 = 9/8 = 1.125 m
∴ Height of embankment = 1.125 m
OR Hemisphere + Cone toy (r = 4 cm, cone height h = 2 cm):
Slant height l = √(r² + h²) = √(16 + 4) = √20 = 2√5 cm
Volume = Volume of cone + Volume of hemisphere = (1/3)πr²h + (2/3)πr³ = πr²/3 × (h + 2r) = (22/7) × 16/3 × (2 + 8) = (22/7) × 16/3 × 10 = 167.62 cm³ (approx)
Total Surface Area = Curved Surface Area of cone + Curved Surface Area of hemisphere = πrl + 2πr² = πr(l + 2r) = (22/7) × 4 × (2√5 + 8) = (88/7) × (4.47 + 8) ≈ (88/7) × 12.47 ≈ 156.8 cm²
Q35. Mean, Median, Mode — Electricity Consumption
| Class | f | Mid-value (x) | cf | f×x |
|---|---|---|---|---|
| 65–85 | 4 | 75 | 4 | 300 |
| 85–105 | 5 | 95 | 9 | 475 |
| 105–125 | 13 | 115 | 22 | 1495 |
| 125–145 | 20 | 135 | 42 | 2700 |
| 145–165 | 14 | 155 | 56 | 2170 |
| 165–185 | 8 | 175 | 64 | 1400 |
| 185–205 | 4 | 195 | 68 | 780 |
| Total | 68 | 9320 |
Mean = Σfx / Σf = 9320/68 = 137.06 units
Median: n/2 = 34 → Cumulative frequency just above 34 is 42 → Median class = 125–145 l = 125, cf = 22, f = 20, h = 20
Median = l + [(n/2 − cf)/f] × h = 125 + [(34 − 22)/20] × 20 = 125 + 12 = 137 units
Mode: Modal class = 125–145 (highest frequency = 20) l = 125, f₁ = 20, f₀ = 13, f₂ = 14, h = 20
Mode = 125 + [(20−13)/(2×20−13−14)] × 20 = 125 + [7/13] × 20 = 125 + 10.77 = 135.77 units
Comparison: Mean ≈ 137.06, Median = 137, Mode ≈ 135.77 All three values are close to each other, indicating the data is nearly symmetrical.
SECTION — E (Case Study Answers)
Q36. Case Study — Real Numbers
(i) Prime Factorisation: 12 = 2² × 3 18 = 2 × 3² 24 = 2³ × 3
(ii) HCF (12, 18, 24): Common factors with lowest powers: 2¹ × 3¹ HCF = 6
(iii) LCM (12, 18, 24): All factors with highest powers: 2³ × 3² = 8 × 9 = 72
Verification: HCF × LCM = 6 × 72 = 432 = 12 × 36 ✓
Q37. Case Study — Coordinate Geometry
Vertices: A(1,2), B(4,6), C(7,2), D(4,−2)
(i) Length of diagonal AC: AC = √[(7−1)² + (2−2)²] = √[36 + 0] = √36 = 6 units
(ii) Length of diagonal BD: BD = √[(4−4)² + (−2−6)²] = √[0 + 64] = √64 = 8 units
(iii) Mid-point of AC: = ((1+7)/2, (2+2)/2) = (4, 2)
Mid-point of BD: = ((4+4)/2, (6+(−2))/2) = (4, 2)
Observation: Both diagonals have the same mid-point (4, 2), meaning the diagonals bisect each other. Since diagonals bisect each other but are not equal (AC = 6 ≠ BD = 8), ABCD is a Rhombus.
Q38. Case Study — Statistics and Probability
Total students = 200
(i) P(student likes Cricket): = 80/200 = 2/5
(ii) P(student likes Football or Basketball): = (50 + 20)/200 = 70/200 = 7/20
(iii) P(student does NOT like Badminton): Students who like Badminton = 40 Students who do NOT like Badminton = 200 − 40 = 160 P(not Badminton) = 160/200 = 4/5
SUMMARY OF ANSWERS — SECTION A
| Q | Ans | Q | Ans |
|---|---|---|---|
| 1 | (a) 4 | 11 | (a) (4,4) |
| 2 | (d) a=0, b=−6 | 12 | (a) 25π cm² |
| 3 | (c) Infinitely many | 13 | (a) 1540 cm³ |
| 4 | (a) 1 and 3/2 | 14 | (c) 1/2 |
| 5 | (b) 48 | 15 | (a) 7 |
| 6 | (b) 4.5 cm | 16 | (b) Perpendicular |
| 7 | (a) 4/5 | 17 | (c) 210 |
| 8 | (d) 1−√3 | 18 | (b) (√3+1)/2 |
| 9 | (c) 30√3 m | 19 | (a) |
| 10 | (d) Both a and b | 20 | (d) |
All the best! Study well and aim for full marks!
Comments
Post a Comment