CBSE Class 10 Maths Standard (Code 041) - 2024 Board Solved Paper

xy

xy²

x³y³

x²y²

Official Verified Answer / Outcome:

B: xy²

Step-by-Step Academic Solution:

Given a = x³ · y² and b = x¹ · y³.
HCF is the product of the smallest power of each common prime factor:
HCF(a, b) = x^(min(3, 1)) · y^(min(2, 3)) = x¹ · y² = xy².

CBSE Marking Guidelines:

Step 1: Identify prime factor powers (½ mark)
Step 2: Select lowest exponent for common bases (½ mark)

4/3

-4/3

2/3

-2/3

Official Verified Answer / Outcome:

A: 4/3

Step-by-Step Academic Solution:

Let P(x) = (k - 1)x² + kx + 1.
Since -3 is a zero, P(-3) = 0:
(k - 1)(-3)² + k(-3) + 1 = 0
=> 9(k - 1) - 3k + 1 = 0
=> 9k - 9 - 3k + 1 = 0
=> 6k - 8 = 0 => 6k = 8 => k = 8/6 = 4/3.

CBSE Marking Guidelines:

Step 1: Equate P(-3) = 0 (½ mark)
Step 2: Solve linear equation for k = 4/3 (½ mark)

1 - √3

1 + √3

0

√3 - 1

Official Verified Answer / Outcome:

A: 1 - √3

Step-by-Step Academic Solution:

Substitute standard values:
sin 30° = 1/2, cos 60° = 1/2 => (sin 30° + cos 60°) = 1/2 + 1/2 = 1.
sin 60° = √3/2, cos 30° = √3/2 => (sin 60° + cos 30°) = √3/2 + √3/2 = √3.
Therefore: 1 - √3.

CBSE Marking Guidelines:

Step 1: Substitute correct trigonometric table values (½ mark)
Step 2: Simplify to 1 - √3 (½ mark)

Official Verified Answer / Outcome:

Proof by Contradiction

Step-by-Step Academic Solution:

1. Let us assume, to the contrary, that √5 is rational.
Hence √5 = a/b, where a and b are co-prime integers and b ≠ 0.
2. Squaring both sides: 5 = a²/b² => a² = 5b².
This means 5 divides a². Therefore, by Fundamental Theorem of Arithmetic, 5 divides a.
3. Let a = 5c for some integer c.
Substituting: (5c)² = 5b² => 25c² = 5b² => b² = 5c².
This implies 5 divides b², so 5 divides b.
4. Hence, 5 is a common factor of both a and b. This contradicts our assumption that a and b are co-prime.
5. Therefore, √5 must be irrational.

CBSE Marking Guidelines:

• Assumption and a² = 5b² expression: 1 mark
• Deducing 5 divides both a and b with final contradiction: 1 mark

Official Verified Answer / Outcome:

Proof LHS = RHS

Step-by-Step Academic Solution:

LHS = [sin θ (1 - 2 sin² θ)] / [cos θ (2 cos² θ - 1)]
Identity: sin² θ + cos² θ = 1 => 1 = sin² θ + cos² θ.
Numerator: 1 - 2 sin² θ = cos² θ - sin² θ.
Denominator: 2 cos² θ - 1 = cos² θ - sin² θ.
LHS = [sin θ · (cos² θ - sin² θ)] / [cos θ · (cos² θ - sin² θ)] = sin θ / cos θ = tan θ = RHS.

CBSE Marking Guidelines:

• Factoring out sin θ and cos θ: 1 mark
• Using trigonometric identity: 1 mark
• Final simplification: 1 mark

Official Verified Answer / Outcome:

Speed of stream = 6 km/h

Step-by-Step Academic Solution:

1. Let the speed of the stream be x km/h (x < 18).
Speed upstream = (18 - x) km/h; speed downstream = (18 + x) km/h.
2. Distance = 24 km.
Time upstream (t₁) = 24 / (18 - x); time downstream (t₂) = 24 / (18 + x).
3. Given: t₁ - t₂ = 1
24/(18 - x) - 24/(18 + x) = 1 => 48x = 324 - x² => x² + 48x - 324 = 0.
4. Factoring: (x - 6)(x + 54) = 0.
Since speed > 0, x = 6 km/h.

CBSE Marking Guidelines:

• Upstream & downstream speed expressions: 2 marks
• Quadratic equation setup: 1½ marks
• Factorization and final unit: 1½ marks

Official Verified Answer / Outcome:

Height = 20√3 m ≈ 34.64 m; Distance from B = 20 m

Step-by-Step Academic Solution:

In △BCD: tan 60° = h/d => h = d√3.
In △ACD: tan 30° = h/(40 + d) => 1/√3 = (d√3)/(40 + d) => 40 + d = 3d => 2d = 40 => d = 20 m.
Height h = 20√3 m.

CBSE Marking Guidelines:

• Right triangle formulation: 1½ marks
• Solving distance d = 20 m: 1 mark
• Calculating height h = 20√3 m: 1½ marks