Class 9 Olympiad Previous Years Question Bank & Solved Papers

IMO Olympiad Year 2024 Class 9 60 Mins · 50 Marks ✓ Verified Solutions

International Mathematics Olympiad (IMO) - Class 9 Solved Paper

Class 9 IMO championship paper covering surd radical decomposition, cyclic quadrilaterals, polynomial remainder theorem, cyclicity of powers, telescoping radical series, symmetric polynomials, coordinate geometry distance, incentre angle theorem, intersecting circle chords, Heron's formula, cone geometry, Archimedes inscribed sphere ratio, and IOQM combinatorics.

Topics Covered: Real Numbers & Surds Polynomials & Remainder Theorem Coordinate Geometry Lines, Angles & Triangles Circles & Chords Heron's Formula & Mensuration Surface Areas & Volumes Number Theory & Combinatorics Achievers HOTS

Question 1 Mathematical Reasoning

1 Mark

If x = 3 + 2√2, what is the value of √x - 1/√x?

A 2 ✓ Correct

B √2

C 2√2

D 4

Concept Tested: Square root of surds and binomial square expansion

Hint: Express x as a perfect square: (√2 + 1)^2.

Detailed Step-by-Step Verified Solution

Step 1: Notice that 3 + 2√2 = 2 + 1 + 2√2 = (√2)^2 + (1)^2 + 2(√2)(1) = (√2 + 1)^2.
Step 2: Therefore, √x = √( (√2 + 1)^2 ) = √2 + 1.
Step 3: Compute 1/√x: 1 / (√2 + 1) = (√2 - 1) / ((√2)^2 - 1^2) = √2 - 1.
Step 4: Calculate √x - 1/√x = (√2 + 1) - (√2 - 1) = √2 + 1 - √2 + 1 = 2.

Official Final Answer: Correct Option (A) : 2

💡 Olympiad Tip & Strategy: Decomposing surds into (a + b)^2 where 2ab = surd radical is a signature Olympiad skill.

Question 2 Mathematical Reasoning

1 Mark

In a cyclic quadrilateral ABCD, if ∠A = (2x + 4)° and ∠C = (3x + 1)°, find the value of ∠C.

A 74°

B 106° ✓ Correct

C 35°

D 108°

Concept Tested: Opposite angles of a cyclic quadrilateral are supplementary (sum 180°)

Hint: ∠A + ∠C = 180°. Solve for x, then substitute in ∠C.

Detailed Step-by-Step Verified Solution

Step 1: In any cyclic quadrilateral, opposite angles sum to 180°: ∠A + ∠C = 180°.
Step 2: (2x + 4) + (3x + 1) = 180° => 5x + 5 = 180°.
Step 3: 5x = 175° => x = 35°.
Step 4: ∠C = 3(35) + 1 = 105 + 1 = 106° (and ∠A = 74°).

Official Final Answer: Correct Option (B) : 106°

💡 Olympiad Tip & Strategy: Always double-check: 74° + 106° = 180°.

Question 3 Achievers Section (HOTS)

3 Marks

If the polynomial P(x) = x⁴ - 2x³ + 3x² - ax + b leaves remainders 5 and 19 when divided by (x - 1) and (x + 1) respectively, what is the value of (a + b)?

A 13 ✓ Correct

B 11

C 7

D 15

Concept Tested: Polynomial Remainder Theorem: P(c) is remainder when divided by (x - c)

Hint: P(1) = 5 and P(-1) = 19. Set up a system of two linear equations in a and b.

Detailed Step-by-Step Verified Solution

Step 1: By Remainder Theorem, P(1) = 5:
1⁴ - 2(1)³ + 3(1)² - a(1) + b = 5 => 1 - 2 + 3 - a + b = 5 => 2 - a + b = 5 => -a + b = 3 (Eq 1).
Step 2: By Remainder Theorem, P(-1) = 19:
(-1)⁴ - 2(-1)³ + 3(-1)² - a(-1) + b = 19 => 1 + 2 + 3 + a + b = 19 => 6 + a + b = 19 => a + b = 13 (Eq 2).
Step 3: Equation 2 gives (a + b) = 13 directly! (Solving further: b = 8, a = 5).

Official Final Answer: Correct Option (A) : 13

💡 Olympiad Tip & Strategy: Using Remainder Theorem directly yields required sums without long division.

Question 4 Logical Reasoning

1 Mark

What is the units digit of the number (7^2024 + 3^2024)?

A 2 ✓ Correct

B 0

C 4

D 8

Concept Tested: Cyclicity of units digits of powers of 7 and 3 (period 4)

Hint: 2024 is a multiple of 4 (2024 mod 4 = 0). Find 7⁴ and 3⁴ units digits.

Detailed Step-by-Step Verified Solution

Step 1: Powers of 7 have cyclicity of 4: 7¹=7, 7²=9, 7³=3, 7⁴=1 (mod 10).
Since 2024 = 4 × 506, units digit of 7^2024 is 1.
Step 2: Powers of 3 also have cyclicity of 4: 3¹=3, 3²=9, 3³=7, 3⁴=1 (mod 10).
Since 2024 = 4 × 506, units digit of 3^2024 is 1.
Step 3: Units digit of the sum = 1 + 1 = 2.

Official Final Answer: Correct Option (A) : 2

💡 Olympiad Tip & Strategy: Most base digits (2, 3, 7, 8) have cyclicity of 4. Divide exponent by 4 to get the remainder.

Question 5 Mathematical Reasoning

1 Mark

Find the simplified value of the telescoping radical sum: 1/(√2 + 1) + 1/(√3 + √2) + 1/(√4 + √3) + ... + 1/(√100 + √99).

A 9 ✓ Correct

B 10

C 8

D √100 - √2

Concept Tested: Rationalization of denominators and telescoping series cancellation

Hint: 1/(√(k+1) + √k) = √(k+1) - √k. All intermediate terms cancel out!

Detailed Step-by-Step Verified Solution

Step 1: Rationalize the general k-th term: 1/(√(k+1) + √k) = (√(k+1) - √k) / ((k+1) - k) = √(k+1) - √k.
Step 2: Write out terms:
(√2 - 1) + (√3 - √2) + (√4 - √3) + ... + (√100 - √99).
Step 3: All terms cancel except -1 and +√100:
Sum = -1 + √100 = -1 + 10 = 9.

Official Final Answer: Correct Option (A) : 9

💡 Olympiad Tip & Strategy: Telescoping series is an elite competition trick: only first and last surviving terms matter.

Question 6 Mathematical Reasoning

1 Mark

If x + y + z = 0, what is the value of (x²/yz) + (y²/zx) + (z²/xy)?

A 3 ✓ Correct

B 1

C 0

D -1

Concept Tested: Conditional algebraic identity: If x + y + z = 0 then x³ + y³ + z³ = 3xyz

Hint: Take common denominator xyz. The numerator is x³ + y³ + z³.

Detailed Step-by-Step Verified Solution

Step 1: Combine into a single fraction: (x³ + y³ + z³) / (xyz).
Step 2: By the classic algebraic identity: x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx).
Step 3: Given x + y + z = 0, the entire RHS becomes 0, so x³ + y³ + z³ = 3xyz.
Step 4: Therefore, (x³ + y³ + z³) / (xyz) = 3xyz / xyz = 3.

Official Final Answer: Correct Option (A) : 3

💡 Olympiad Tip & Strategy: Recognizing conditional factorizations saves lengthy expansions.

Question 7 Mathematical Reasoning

1 Mark

Find a point on the x-axis which is equidistant from the points A(2, -5) and B(-2, 9).

A (-7, 0) ✓ Correct

B (7, 0)

C (0, -7)

D (-5, 0)

Concept Tested: Coordinate geometry distance formula and points on the x-axis (x, 0)

Hint: Let P be (x, 0). Equate PA² = PB².

Detailed Step-by-Step Verified Solution

Step 1: Any point on the x-axis has coordinates P(x, 0).
Step 2: Distance PA² = (x - 2)² + (0 - (-5))² = (x - 2)² + 25.
Step 3: Distance PB² = (x - (-2))² + (0 - 9)² = (x + 2)² + 81.
Step 4: Equate PA² = PB²: x² - 4x + 4 + 25 = x² + 4x + 4 + 81.
-4x + 29 = 4x + 85 => 8x = -56 => x = -7.
Step 5: The required point is (-7, 0).

Official Final Answer: Correct Option (A) : (-7, 0)

💡 Olympiad Tip & Strategy: Always use squared distances to avoid carrying square root signs throughout algebra.

Question 8 Mathematical Reasoning

1 Mark

In triangle ABC, the internal bisectors of ∠B and ∠C intersect at the incentre I. If ∠BAC = 70°, what is the measure of ∠BIC?

A 125° ✓ Correct

B 110°

C 135°

D 140°

Concept Tested: Incentre angle property: ∠BIC = 90° + ∠A / 2

Hint: In ΔBIC, ∠BIC = 180° - (∠B/2 + ∠C/2) = 180° - (180° - ∠A)/2 = 90° + ∠A/2.

Detailed Step-by-Step Verified Solution

Step 1: In ΔABC, ∠A + ∠B + ∠C = 180° => ∠B + ∠C = 180° - ∠A.
Step 2: In ΔBIC, ∠BIC + ∠B/2 + ∠C/2 = 180°.
∠BIC = 180° - (∠B + ∠C)/2 = 180° - (180° - ∠A)/2 = 90° + ∠A/2.
Step 3: Substitute ∠A = 70°: ∠BIC = 90° + 70°/2 = 90° + 35° = 125°.

Official Final Answer: Correct Option (A) : 125°

💡 Olympiad Tip & Strategy: Olympiad Incentre Formula: ∠BIC = 90° + A/2; Excentre angle = 90° - A/2!

Question 9 Mathematical Reasoning

1 Mark

Two intersecting circles of equal radii 5 cm have their centres 8 cm apart. What is the length of their common chord?

A 6 cm ✓ Correct

B 8 cm

C 7 cm

D 4 cm

Concept Tested: Properties of common chords of intersecting circles: line of centres bisects common chord at 90°

Hint: Line of centres bisects chord AB at midpoint M. OM = 8/2 = 4 cm. Use Pythagoras on ΔOMA.

Detailed Step-by-Step Verified Solution

Step 1: Let centres be O1 and O2 with O1O2 = 8 cm. Radius r = 5 cm.
Step 2: The line connecting centres is the perpendicular bisector of the common chord AB at point M.
Step 3: Distance from centre to chord M = 8 ÷ 2 = 4 cm.
Step 4: In right triangle O1MA: AM = √(r² - OM²) = √(5² - 4²) = √(25 - 16) = √9 = 3 cm.
Step 5: Total length of common chord AB = 2 × AM = 2 × 3 = 6 cm.

Official Final Answer: Correct Option (A) : 6 cm

💡 Olympiad Tip & Strategy: The 3-4-5 right triangle appears frequently in circle intersecting chord problems.

Question 10 Everyday Mathematics / Science

1 Mark

The sides of a triangular park are 50 m, 65 m, and 65 m. Find the total area of the park using Heron's formula.

A 1500 m² ✓ Correct

B 1625 m²

C 1400 m²

D 1750 m²

Concept Tested: Heron's Formula for area of triangle: Area = √[s(s - a)(s - b)(s - c)]

Hint: Semi-perimeter s = (50 + 65 + 65)/2 = 90 m.

Detailed Step-by-Step Verified Solution

Step 1: Calculate semi-perimeter s = (50 + 65 + 65) / 2 = 180 / 2 = 90 m.
Step 2: Compute differences: (s - a) = 90 - 50 = 40; (s - b) = 90 - 65 = 25; (s - c) = 90 - 65 = 25.
Step 3: Area = √(90 × 40 × 25 × 25) = √(3600 × 625) = 60 × 25 = 1500 m².

Official Final Answer: Correct Option (A) : 1500 m²

💡 Olympiad Tip & Strategy: Notice the isosceles shortcut: Altitude to 50 m base is √(65² - 25²) = √(4225 - 625) = √3600 = 60 m! Area = 1/2 × 50 × 60 = 1500 m².

Question 11 Everyday Mathematics / Science

1 Mark

A conical tent of radius 7 m and vertical height 24 m is made of canvas cloth. Find the slant height and the area of canvas needed to pitch the tent. (Take π = 22/7)

A Slant height = 25 m; Area = 550 m² ✓ Correct

B Slant height = 25 m; Area = 616 m²

C Slant height = 26 m; Area = 572 m²

D Slant height = 25 m; Area = 528 m²

Concept Tested: Cone slant height l = √(r² + h²) and curved surface area CSA = πrl

Hint: 7-24-25 is a Pythagorean triplet. CSA of cone = π × r × l.

Detailed Step-by-Step Verified Solution

Step 1: Slant height l = √(r² + h²) = √(7² + 24²) = √(49 + 576) = √625 = 25 m.
Step 2: Area of canvas required = Curved Surface Area = π × r × l.
Step 3: CSA = (22/7) × 7 × 25 = 22 × 25 = 550 m².

Official Final Answer: Correct Option (A) : Slant height = 25 m; Area = 550 m²

💡 Olympiad Tip & Strategy: The base of a tent is open ground, so do not include the base circle πr²!

Question 12 Achievers Section (HOTS)

3 Marks

A solid sphere of radius R is snugly inscribed inside a right circular cylinder such that the sphere touches the top, bottom, and curved surface of the cylinder. What is the ratio of the volume of the sphere to the volume of the cylinder?

A 2 : 3 ✓ Correct

B 3 : 4

C 1 : 2

D 4 : 5

Concept Tested: Archimedes famous sphere-cylinder volume theorem

Hint: Cylinder has radius R and height H = 2R. Compute Volume_sphere / Volume_cylinder.

Detailed Step-by-Step Verified Solution

Step 1: The inscribed sphere touches both flat faces, so cylinder height H = 2R.
The cylinder radius is also R.
Step 2: Volume of sphere = (4/3) π R³.
Step 3: Volume of cylinder = π R² H = π R² (2R) = 2 π R³.
Step 4: Ratio = [(4/3) π R³] / [2 π R³] = (4/3) ÷ 2 = 2/3 = 2 : 3.

Official Final Answer: Correct Option (A) : 2 : 3

💡 Olympiad Tip & Strategy: Archimedes was so proud of this 2:3 ratio that he requested a sphere inscribed in a cylinder engraved on his tombstone!

Question 13 Achievers Section (HOTS)

3 Marks

How many 4-digit numbers can be formed using the digits {1, 2, 3, 4, 5} without repetition such that the number is strictly divisible by 4?

A 24 ✓ Correct

B 18

C 12

D 30

Concept Tested: Divisibility rule for 4 (last two digits must be divisible by 4) and permutations

Hint: Find all 2-digit pairs divisible by 4 from {1, 2, 3, 4, 5}: 12, 24, 32, 52 (4 pairs). First two digits have 3 × 2 options.

Detailed Step-by-Step Verified Solution

Step 1: A number is divisible by 4 if its last two digits form a multiple of 4.
Step 2: Permutations of two distinct digits from {1, 2, 3, 4, 5} divisible by 4:
Valid endings: 12, 24, 32, 52 (Note: 44 is excluded as repetition is not allowed).
There are exactly 4 valid endings.
Step 3: For each ending, 2 digits are used up, leaving 3 available digits.
The thousands and hundreds places can be filled in 3 × 2 = 6 ways.
Step 4: Total valid 4-digit numbers = 4 endings × 6 ways = 24 numbers.

Official Final Answer: Correct Option (A) : 24

💡 Olympiad Tip & Strategy: Always isolate the condition on the trailing digits first in divisibility combinatorics.

NSO Olympiad Year 2024 Class 9 60 Mins · 50 Marks ✓ Verified Solutions

National Science Olympiad (NSO) - Class 9 Solved Paper

Class 9 NSO paper covering conservation of mechanical energy, mole concept and atomicity, graphical analysis of motion (v-t graphs), Newton's second law with combined masses, gravitation and g variation, Archimedes floatation principle, mechanical power calculations, wave velocity and sound propagation, lysosome cytology, plant tissues (collenchyma), stoichiometry of thermal decomposition, and SONAR ultrasonic echo calculations.

Topics Covered: Motion & Kinematics Graphs Force & Laws of Motion Gravitation & Floatation Work, Energy & Power Sound & Ultrasonics Atoms, Molecules & Mole Concept Cell Structure & Organelles Plant Tissues & Histology Chemical Reactions & Stoichiometry Achievers HOTS

Question 1 Scientific Reasoning

1 Mark

A ball of mass 200 g is thrown vertically upward with an initial velocity of 20 m/s. What is its kinetic energy when it reaches half of its maximum height? (Take g = 10 m/s²)

A 40 J

B 20 J ✓ Correct

C 10 J

D 5 J

Concept Tested: Work-Energy Theorem and Conservation of Mechanical Energy

Hint: Total mechanical energy = Initial KE = (1/2) m v^2. At half height, half the energy is PE and half is KE.

Detailed Step-by-Step Verified Solution

Step 1: Convert mass to SI units: m = 200 g = 0.2 kg.
Step 2: Total initial kinetic energy at ground: KE_initial = (1/2) × 0.2 × (20)^2 = 0.1 × 400 = 40 J.
Step 3: At maximum height H, all 40 J is converted into potential energy (PE = 40 J).
Step 4: At half maximum height (H/2), potential energy is PE = 40 ÷ 2 = 20 J.
Step 5: By Conservation of Mechanical Energy (E = KE + PE = 40 J): KE = 40 - 20 = 20 J.

Official Final Answer: Correct Option (B) : 20 J

💡 Olympiad Tip & Strategy: Using energy conservation avoids calculating the numerical height H altogether!

Question 2 Achievers Section (HOTS)

3 Marks

Calculate the number of moles and total atoms present in 54 g of pure water (H2O). (Given: Molar mass of H = 1 g/mol, O = 16 g/mol, NA = 6.022 × 10^23)

A 3 moles; 1.8066 × 10^24 atoms

B 3 moles; 5.42 × 10^24 atoms ✓ Correct

C 2 moles; 3.61 × 10^24 atoms

D 4 moles; 7.22 × 10^24 atoms

Concept Tested: Mole concept and atomicity of polyatomic molecules (H2O has 3 atoms per molecule)

Hint: Number of moles = mass ÷ molar mass. Total atoms = moles × NA × atomicity (3).

Detailed Step-by-Step Verified Solution

Step 1: Molar mass of H2O = 2(1) + 16 = 18 g/mol.
Step 2: Number of moles n = 54 g ÷ 18 g/mol = 3 moles of H2O molecules.
Step 3: Total molecules = 3 × NA = 3 × 6.022 × 10^23 = 1.8066 × 10^24 molecules.
Step 4: Each molecule of H2O contains 3 atoms (2 Hydrogen + 1 Oxygen).
Total atoms = 3 × 1.8066 × 10^24 = 5.4198 × 10^24 ≈ 5.42 × 10^24 atoms.

Official Final Answer: Correct Option (B) : 3 moles; 5.42 × 10^24 atoms

💡 Olympiad Tip & Strategy: Notice the question asks for total *atoms*, not just total molecules!

Question 3 Scientific Reasoning

1 Mark

In a velocity-time (v-t) graph of a moving object, the area enclosed between the curve and the time axis represents:

A Displacement of the body ✓ Correct

B Acceleration of the body

C Speed of the body

D Rate of change of momentum

Concept Tested: Graphical analysis of motion: Area under v-t graph = Displacement

Hint: Velocity × Time = [m/s] × [s] = meters (displacement).

Detailed Step-by-Step Verified Solution

Step 1: Velocity is rate of change of displacement: v = ds/dt => ds = v·dt.
Step 2: The integral or geometric area under the v-t graph represents the product of velocity and time.
Step 3: Hence, the enclosed area equals the magnitude of displacement traversed by the object.

Official Final Answer: Correct Option (A) : Displacement of the body

💡 Olympiad Tip & Strategy: Remember: The *slope* of a v-t graph represents acceleration; the *area* represents displacement.

Question 4 Scientific Reasoning

1 Mark

A constant horizontal force of 6 N produces an acceleration of 2 m/s² on a body of mass m1, and 6 m/s² on a body of mass m2. What acceleration would this same force produce if both masses were tied together?

A 1.5 m/s² ✓ Correct

B 2.0 m/s²

C 3.0 m/s²

D 4.0 m/s²

Concept Tested: Newton's Second Law: F = ma and equivalent combined mass

Hint: m1 = F/a1 = 6/2 = 3 kg. m2 = F/a2 = 6/6 = 1 kg. Combined mass = 4 kg.

Detailed Step-by-Step Verified Solution

Step 1: From F = ma, mass m1 = F / a1 = 6 N / 2 m/s² = 3 kg.
Step 2: Mass m2 = F / a2 = 6 N / 6 m/s² = 1 kg.
Step 3: Total combined mass M = m1 + m2 = 3 + 1 = 4 kg.
Step 4: Acceleration produced by same force: a_combined = F / M = 6 N / 4 kg = 1.5 m/s².

Official Final Answer: Correct Option (A) : 1.5 m/s²

💡 Olympiad Tip & Strategy: Shortcut formula: a = (a1 × a2) / (a1 + a2) = (2 × 6) / (2 + 6) = 12 / 8 = 1.5 m/s².

Question 5 Scientific Reasoning

1 Mark

If the radius of the Earth were to shrink by 1% while its mass remained constant, how would the acceleration due to gravity (g) at the surface change?

A Increases by approximately 2% ✓ Correct

B Decreases by approximately 2%

C Increases by 1%

D Remains completely unchanged

Concept Tested: Universal Gravitation and fractional differentiation of g = GM/R²

Hint: g ∝ 1/R². If R decreases by 1%, g increases by 2(1%) = 2%.

Detailed Step-by-Step Verified Solution

Step 1: Formula for surface gravity: g = G·M / R².
Step 2: Using fractional error/logarithmic differentiation: Δg/g = -2 (ΔR/R).
Step 3: Since R shrinks by 1% (ΔR/R = -1%):
Δg/g = -2 (-1%) = +2%.
Step 4: Therefore, acceleration due to gravity increases by approximately 2%.

Official Final Answer: Correct Option (A) : Increases by approximately 2%

💡 Olympiad Tip & Strategy: Power rule: For y = x^n, fractional change in y is n times fractional change in x.

Question 6 Scientific Reasoning

1 Mark

A block of wood of density 600 kg/m³ floats in pure water (density 1000 kg/m³). What fraction of the block's total volume remains submerged underwater?

A 0.6 (or 60%) ✓ Correct

B 0.4 (or 40%)

C 0.75 (or 75%)

D 0.5 (or 50%)

Concept Tested: Archimedes principle and Law of Floatation: V_sub / V_total = ρ_body / ρ_liquid

Hint: Weight of floating body = Weight of displaced water.

Detailed Step-by-Step Verified Solution

Step 1: By the Law of Floatation, Weight of floating body = Buoyant force = Weight of displaced liquid.
Step 2: V_total × ρ_body × g = V_submerged × ρ_liquid × g.
Step 3: V_submerged / V_total = ρ_body / ρ_liquid = 600 / 1000 = 0.6 = 60%.

Official Final Answer: Correct Option (A) : 0.6 (or 60%)

💡 Olympiad Tip & Strategy: The fraction *above* water is (1 - 0.6) = 0.4 (40%). Read carefully whether submerged or exposed is asked!

Question 7 Scientific Reasoning

1 Mark

An electric crane lifts a load of 500 kg vertically upward through a height of 20 m in 10 seconds. What is the useful mechanical power delivered by the crane? (Take g = 10 m/s²)

A 10 kW ✓ Correct

B 5 kW

C 20 kW

D 1 kW

Concept Tested: Work, Energy and Power: Power = Work Done ÷ Time = (mgh) / t

Hint: Work done = m × g × h = 500 × 10 × 20 = 100,000 J. Divide by 10 s.

Detailed Step-by-Step Verified Solution

Step 1: Work done against gravity W = m·g·h = 500 kg × 10 m/s² × 20 m = 100,000 Joules (100 kJ).
Step 2: Power delivered P = Work Done / Time taken = 100,000 J / 10 s = 10,000 Watts.
Step 3: Convert to kilowatts: 10,000 W = 10 kW.

Official Final Answer: Correct Option (A) : 10 kW

💡 Olympiad Tip & Strategy: 1 Watt = 1 Joule/second; 1 Horsepower (hp) = 746 Watts.

Question 8 Scientific Reasoning

1 Mark

A sound wave has a frequency of 2 kHz and wavelength of 35 cm. How long will it take to travel a distance of 1.4 km through this medium?

A 2.0 seconds ✓ Correct

B 1.5 seconds

C 3.0 seconds

D 0.5 seconds

Concept Tested: Wave equation v = ν λ and kinematics travel time t = d / v

Hint: Convert frequency to Hz (2000 Hz) and wavelength to meters (0.35 m). Speed v = 2000 × 0.35 = 700 m/s.

Detailed Step-by-Step Verified Solution

Step 1: Convert into SI units: Frequency ν = 2 kHz = 2000 Hz; Wavelength λ = 35 cm = 0.35 m.
Step 2: Calculate wave velocity: v = ν × λ = 2000 × 0.35 = 700 m/s.
Step 3: Distance to travel d = 1.4 km = 1400 m.
Step 4: Time taken t = Distance / Velocity = 1400 m / 700 m/s = 2.0 seconds.

Official Final Answer: Correct Option (A) : 2.0 seconds

💡 Olympiad Tip & Strategy: Always convert kHz to Hz and cm to m before multiplying!

Question 9 Scientific Reasoning

1 Mark

Which cell organelle contains powerful digestive hydrolytic enzymes capable of digesting the cell itself under cellular damage, earning the moniker "suicide bags"?

A Lysosomes ✓ Correct

B Mitochondria

C Ribosomes

D Golgi Apparatus

Concept Tested: Cell biology: Function and pathology of lysosomes and autolysis

Hint: Lysosomes contain membrane-bound hydrolytic enzymes synthesized by rough ER.

Detailed Step-by-Step Verified Solution

Step 1: Lysosomes are membrane-enclosed vesicles containing acidic hydrolytic digestive enzymes.
Step 2: When a cell gets damaged or undergoes cellular starvation/stress, lysosomes burst open.
Step 3: The released enzymes digest the cell's own components (autolysis), hence called "suicide bags".

Official Final Answer: Correct Option (A) : Lysosomes

💡 Olympiad Tip & Strategy: Mitochondria are the "powerhouse", and Ribosomes are the "protein factories".

Question 10 Everyday Science

1 Mark

Which plant tissue provides flexible mechanical support allowing bending of tendrils and climbing stems without breaking?

A Collenchyma ✓ Correct

B Sclerenchyma

C Parenchyma

D Aerenchyma

Concept Tested: Plant histology: Collenchyma cells with pectin thickenings at corners

Hint: Collenchyma provides flexibility with strength; Sclerenchyma provides rigid hardness (dead cells).

Detailed Step-by-Step Verified Solution

Step 1: Collenchyma consists of living elongated cells with irregularly thickened walls (thickened with pectin and cellulose at the corners).
Step 2: It provides tensile strength combined with high elasticity and flexibility.
Step 3: This allows aerial plant parts like petiole, pedicel, and tendrils to bend freely in strong winds without snapping.

Official Final Answer: Correct Option (A) : Collenchyma

💡 Olympiad Tip & Strategy: Husk of coconut is made of dead Sclerenchyma tissue with lignin!

Question 11 Achievers Section (HOTS)

3 Marks

When 10.0 g of pure calcium carbonate (CaCO3) is completely thermally decomposed, 5.6 g of solid calcium oxide (CaO) is formed. According to the Law of Conservation of Mass, what mass of carbon dioxide (CO2) gas is liberated, and how many moles does it correspond to?

A 4.4 g of CO2; 0.1 mole ✓ Correct

B 4.4 g of CO2; 0.2 mole

C 5.0 g of CO2; 0.1 mole

D 3.6 g of CO2; 0.08 mole

Concept Tested: Law of Conservation of Mass and molar mass of CO2 (44 g/mol)

Hint: Mass of reactants = Mass of products. 10.0 g = 5.6 g + Mass(CO2). Moles = mass / 44.

Detailed Step-by-Step Verified Solution

Step 1: Reaction: CaCO3 (s) -> CaO (s) + CO2 (g).
Step 2: By Law of Conservation of Mass: Mass of CaCO3 = Mass of CaO + Mass of CO2.
10.0 g = 5.6 g + Mass(CO2) => Mass of CO2 = 10.0 - 5.6 = 4.4 g.
Step 3: Molar mass of CO2 = 12 + 2(16) = 44 g/mol.
Step 4: Moles of CO2 = Mass / Molar mass = 4.4 g / 44 g/mol = 0.1 mole.

Official Final Answer: Correct Option (A) : 4.4 g of CO2; 0.1 mole

💡 Olympiad Tip & Strategy: At STP, 0.1 mole of any ideal gas occupies exactly 2.24 litres (2240 mL) of volume.

Question 12 Achievers Section (HOTS)

3 Marks

A sound navigation system (SONAR) on a research ship sends out an ultrasonic pulse towards the ocean seabed. The reflected echo is detected after 3.2 seconds. If the speed of sound in seawater is 1500 m/s, what is the depth of the ocean bed at that location?

A 2400 m ✓ Correct

B 4800 m

C 1200 m

D 3000 m

Concept Tested: Echo and SONAR depth formula: 2d = v × t => d = (v × t) / 2

Hint: The sound travels down to the seabed and back up to the detector (round trip = 2d).

Detailed Step-by-Step Verified Solution

Step 1: Total distance covered by sound pulse during round trip = 2 × Depth (2d).
Step 2: Formula: 2d = Speed of sound (v) × Echo time (t).
Step 3: 2d = 1500 m/s × 3.2 s = 4800 m.
Step 4: Depth d = 4800 ÷ 2 = 2400 m (2.4 km).

Official Final Answer: Correct Option (A) : 2400 m

💡 Olympiad Tip & Strategy: Always remember to divide by 2 for echo and SONAR reflections!