RBSE Class 8th 2026 HALF-YEARLY-MATHEMATICS-011225 Previous Year Papers

Class: 8th
Subject: Maths
Time: 2 Hr 30 Min.   Maximum Marks: 50

Note:

1. All questions are compulsory.
2. Marks are indicated against each question, attempt every part carefully.

Q.1 Choose the correct option. (1 × 8 = 8 Marks)

1. The solution of the equation x - 2 = 7 is:

   • (A) 5
   • (B) 7
   • (C) 9
   • (D) 14

   Explanation: x - 2 = 7 ⇒ x = 7 + 2 = 9.

2. A planar figure with four sides is called a:

   • (A) diagonal
   • (B) quadrilateral
   • (C) triangle
   • (D) circle

   Explanation: A quadrilateral is a polygon with four sides.

3. The complete angle at the centre of a circle is:

   • (A) 90°
   • (B) 360°
   • (C) 480°
   • (D) 200°

   Explanation: A full circle measures 360° at the centre.

4. The square root of 100 is:

   • (A) 20
   • (B) 10
   • (C) 100
   • (D) 1000

   Explanation: √100 = 10 because 10 × 10 = 100.

5. The value of 8² will be:

   • (A) 64
   • (B) 240
   • (C) 26
   • (D) 522

   Explanation: 8² = 8 × 8 = 64.

6. The sum of 7x, 10x, and 12x is:

   • (A) 29x
   • (B) 29x²
   • (C) 29
   • (D) None of these

   Explanation: 7x + 10x + 12x = (7+10+12)x = 29x.

7. (6x + 2) squared is:

   • (A) 36x² + 4
   • (B) 36x² + 6x + 4
   • (C) 36x² + 24x + 4
   • (D) None

   Explanation: (6x + 2)² = (6x)² + 2(6x)(2) + 2² = 36x² + 24x + 4.

8. Find a common factor of 36 and 2x:

   • (A) 2
   • (B) 2x
   • (C) 3
   • (D) 36

   Explanation: 36 = 2 × 18 and 2x = 2 × x, so common factor is 2.

Q.2 Fill in the blanks. (1 × 8 = 8 Marks)

1. Value of (3² + 4² + 5²) is 50.

   Solution: 3² = 9, 4² = 16, 5² = 25. Sum = 9 + 16 + 25 = 50.

2. The exponential form of z × z × z × z is z⁴.

   Solution: z multiplied by itself 4 times is written as z⁴.

3. The sum of (8x² + 2x) and (4x + 2) is 8x² + 6x + 2.

   Solution: (8x² + 2x) + (4x + 2) = 8x² + (2x + 4x) + 2 = 8x² + 6x + 2.

4. Cubes of even numbers are even.

   Solution: Even numbers are divisible by 2, so their cubes are also divisible by 2, hence even.

5. 2⁴ = 4² = 16.

   Solution: 2⁴ = 2 × 2 × 2 × 2 = 16 and 4² = 4 × 4 = 16.

6. Probability of an event = (Number of outcomes that make up the event) / (Total number of possible outcomes).

   Solution: Probability = (Favourable outcomes) / (Total outcomes).

7. -1 is a rational number greater than -2.

   Solution: -1 is a rational number (since it can be written as -1/1) and -1 > -2.

Question (i) to (v) – True/False

1. Rational numbers are of the form p/q where q ≠ 0. True
2. There is no perfect cube that ends in 8. False (e.g., 2³ = 8, 12³ = 1728)
3. The sum of -5x and 5x is 10x. False (sum is 0)
4. The product of (x + 7) and (y – 5) is xy – 5x – 7y – 35. False (correct product: xy – 5x + 7y – 35)
5. aᵐ × aⁿ = aᵐ⁺ⁿ. True

Q.4 Find the value of n for which 5ⁿ + 5³ = 5³ (2 Marks)

Q.5 Find the product of the following pairs of monomials (2 Marks)

1. (–3p) × (–3p) = 9p²
2. (–a²) × (7pq) = –7a²pq
3. (4) × (0) = 0

Q.6 Explain which property allows you to calculate (6 × ⅓) × 3 as 6 × (⅓ × 3) (2 Marks)

Q.7 Solve the following equations and check your answer (4 Marks)

1. 3m = 5m – 8
   3m – 5m = –8 → –2m = –8 → m = 4
   Check: LHS = 3×4 = 12; RHS = 5×4 – 8 = 20 – 8 = 12. ✓
2. 2y + 3 = 7
   2y = 7 – 3 = 4 → y = 2
   Check: LHS = 2×2 + 3 = 4 + 3 = 7 = RHS. ✓

Q.8 How many sides does a regular polygon have if one exterior angle measures 24°? (2 Marks)

Q.9 When a die is thrown, write the outcome of each of the following events (3 Marks)

1. A number greater than 5: {6}
2. Not a prime number: {1, 4, 6} (since prime numbers on a die are 2, 3, 5)

Q.10 Find the square roots of 200 and 69 by the division method (2 Marks)

Q.11 Find the length of the side of a square whose area is 44 m² (2 Marks)

Q.12 Find the cube root of each of the following numbers by the prime factorization method (4 Marks)

1. 475376: Prime factorization: 475376 = 2⁴ × 11 × 37 × 73 (not a perfect cube; cube root ≈ 78.0)
2. 64: 64 = 2⁶ → cube root = 2² = 4
3. 522: 522 = 2 × 3² × 29 (not a perfect cube; cube root ≈ 8.05)

Q.13 Complete the table (5 Marks)

Q.14 Suppose 2 kg of sugar contains 9 × 10⁶ crystals. How many sugar crystals will the following sugar contain? (3 Marks)

1. 5 kg: Crystals in 1 kg = (9 × 10⁶)/2 = 4.5 × 10⁶. So in 5 kg = 5 × 4.5 × 10⁶ = 22.5 × 10⁶ = 2.25 × 10⁷ crystals.
2. 1.2 kg: Crystals = 1.2 × 4.5 × 10⁶ = 5.4 × 10⁶ crystals.

Q.15 Find the values of unknown x, y, z in the following parallelograms (3 Marks)