Class 10 Maths Half Yearly 2025: Key Topics Solved Step-by-Step

Real Numbers: Essential Problem Solving

Prime factorization is foundational for HCF problems. Consider finding the HCF of 96 and 404. Factorize 96 as (2^5 \times 3) and 404 as (2^2 \times 101). The common prime factor is (2^2), so HCF is 4. This method is reliable for any integer pair—systematically list factors to avoid errors.

When given the product of two numbers (12,960) and their HCF (18), use the formula:
[
\text{Product} = \text{HCF} \times \text{LCM}
]
Substitute values: (12,960 = 18 \times \text{LCM}), so (\text{LCM} = 720). This relationship is non-negotiable in number theory.

Proving (\sqrt{5}) is irrational requires contradiction. Assume it's rational ((\sqrt{5} = p/q) where (p, q) are coprime). Then (5 = p^2/q^2), so (p^2 = 5q^2). This implies 5 divides (p), so (p = 5k). Substituting: (25k^2 = 5q^2) → (q^2 = 5k^2). Thus 5 divides (q), contradicting coprimality. This proof structure applies to any prime number’s square root.

Polynomials: Roots and Relationships

For the quadratic (kx^2 + 5x + 3 = 0), sum of roots is (-5/k) and product is (3/k). If sum equals product:
[
-\frac{5}{k} = \frac{3}{k} \quad \Rightarrow \quad -5 = 3
]
This inconsistency arises only if (k
eq 0). Correctly: (-\frac{5}{k} = 3) → (k = -\frac{5}{3}). Always verify equation setup to avoid sign errors.

Given (3x^2 - 5x + 9) with roots (\alpha, \beta):
[
\alpha + \beta = -\frac{b}{a} = \frac{5}{3}, \quad \alpha\beta = \frac{c}{a} = 3
]
These values help reconstruct equations—critical for word problems.

Linear Equations: Practical Applications

Solve: "5 apples + 3 oranges = ₹35; 2 apples + 4 oranges = ₹28." Let apple cost = (x), orange cost = (y):
[
5x + 3y = 35 \quad \text{(1)} \
2x + 4y = 28 \quad \text{(2)}
]
Multiply (1) by 4 and (2) by 3:
[
20x + 12y = 140 \
6x + 12y = 84
]
Subtract: (14x = 56) → (x = 4). Substitute into (1): (20 + 3y = 35) → (y = 5). Graphically, lines intersect at (4,5)—plot both equations to visualize.

"Two numbers differ by 26; one is triple the other." Let smaller number = (x), larger = (3x):
[
3x - x = 26 \quad \Rightarrow \quad x = 13, \quad 3x = 39
]
Define variables clearly to prevent misassignment.

Quadratic Equations: Solving Techniques

Solve (\frac{x-1}{x-2} + \frac{x-3}{x-4} = \frac{10}{3}). Combine terms:
[
\frac{(x-1)(x-4) + (x-3)(x-2)}{(x-2)(x-4)} = \frac{10}{3}
]
Numerator: (x^2 - 5x + 4 + x^2 - 5x + 6 = 2x^2 - 10x + 10). Denominator: (x^2 - 6x + 8). Equation becomes:
[
\frac{2x^2 - 10x + 10}{x^2 - 6x + 8} = \frac{10}{3}
]
Cross-multiply: (3(2x^2 - 10x + 10) = 10(x^2 - 6x + 8)) → (6x^2 - 30x + 30 = 10x^2 - 60x + 80). Simplify:
[
-4x^2 + 30x - 50 = 0 \quad \Rightarrow \quad 2x^2 - 15x + 25 = 0
]
Factors: ((2x - 5)(x - 5) = 0) → (x = 2.5) or (x = 5). Exclude invalid domains where denominator is zero ((x
eq 2, 4)).

Find consecutive integers summing to 365: Let numbers be (n, n+1). Then:
[
n^2 + (n+1)^2 = 365 \quad \Rightarrow \quad 2n^2 + 2n - 364 = 0 \quad \Rightarrow \quad n^2 + n - 182 = 0
]
Factors: ((n + 14)(n - 13) = 0) → (n = 13) (since positive). Numbers: 13, 14. Check solutions: (13^2 + 14^2 = 169 + 196 = 365).

Arithmetic Progressions: Terms and Sums

For AP: 3, 7, 11, ... First term (a = 3), common difference (d = 4). 11th term:
[
T_{11} = a + 10d = 3 + 40 = 43
]
Sum of first 200 positive integers:
[
S_n = \frac{n(n+1)}{2} = \frac{200 \times 201}{2} = 20,100
]
If second term = 24, third term = 28:
[
a + d = 24, \quad a + 2d = 28 \quad \Rightarrow \quad d = 4, \quad a = 20
]
Sum of 61 terms:
[
S_{61} = \frac{61}{2} \left[ 2 \times 20 + 60 \times 4 \right] = \frac{61}{2} \times 280 = 61 \times 140 = 8,540
]
Use standard formulas—they’re efficient for exams.

Triangles: Similarity Proofs

Two triangles are similar if:

• Corresponding angles are equal, or
• Corresponding sides are proportional (SSS), or
• Two sides proportional and included angle equal (SAS).
  Proof for SAS similarity: If (\triangle ABC) and (\triangle DEF) have (\angle A = \angle D) and (\frac{AB}{DE} = \frac{AC}{DF}), then by definition, they are similar. NCERT Theorem 6.5 validates this.

Coordinate Geometry: Key Concepts

Prove collinearity of A(2,3), B(4,k), C(6,-3). Slopes must be equal:
[
\text{Slope}{AB} = \frac{k-3}{4-2} = \frac{k-3}{2}, \quad \text{Slope}{BC} = \frac{-3 - k}{6-4} = \frac{-3-k}{2}
]
Set equal: (\frac{k-3}{2} = \frac{-3-k}{2}) → (k - 3 = -3 - k) → (2k = 0) → (k = 0).

Midpoint of segment joining (0,0) and (5,-3):
[
\left( \frac{0+5}{2}, \frac{0 + (-3)}{2} \right) = (2.5, -1.5)
]
Plot points to visualize geometric relationships.

Exam Success Toolkit

Actionable Checklist:

1. Practice prime factorization for 10+ numbers.
2. Solve 3 quadratic equations using different methods.
3. Derive 2 AP sums using (S_n = \frac{n}{2} [2a + (n-1)d]).
4. Verify triangle similarity for 2 figure pairs.
5. Calculate slopes for 5 coordinate sets.

Recommended Resources:

• NCERT Textbook: Primary authority for theory and problem patterns.
• GeoGebra: Visualize graphs and geometric proofs.
• RD Sharma Solutions: For advanced practice.

Final Insight: Consistent practice of these topics builds exam confidence. Which problem type challenges you most? Share in the comments—we’ll address it in Part 2!