Extra Questions: Class 8 Maths Chapter 1 A Square and A Cube

Practice questions beyond the textbook exercises, designed to test deeper understanding of squares, cubes and their roots for Class 8 Maths Chapter 1: A Square and A Cube. These Class 8 Mathematics Chapter 1 important questions are handy for last-minute exam practice.

Extra Questions: Class 8 Maths Chapter 1 A Square and A Cube

1. 1 (Assertion-Reason). Assertion: 1352 is not a perfect square. Reason: A perfect square never ends in 2, 3, 7 or 8.
   Solution: Both the assertion and reason are true, and the reason correctly explains the assertion — 1352 ends in 2, so it cannot be a perfect square.
2. 2 (Numerical). Find the smallest number by which 3675 must be divided so that the quotient is a perfect square. Also find the square root of the quotient.
   Solution: 3675 = 3×5²×7². The lone 3 must be removed, so divide by 3; the quotient is 1225, and √1225 = 35.
3. 3 (Numerical). Find the smallest number by which 2560 must be divided so that the quotient is a perfect cube. Also find the cube root of the quotient.
   Solution: 2560 = 2⁹×5¹. For a perfect cube, every prime’s exponent must be a multiple of 3. The exponent of 2 (nine) already qualifies, but the exponent of 5 (one) does not, so divide by 5. The quotient is 512 = 2⁹ = 8³, so ∛512 = 8.
4. 4 (Applied). A gardener has 5476 saplings and wants to plant them in a square-shaped grid with none left over. Is this possible? If yes, how many rows?
   Solution: 5476 = 74², a perfect square, so yes — a 74×74 grid uses exactly 5476 saplings with 74 rows.
5. 5 (Assertion-Reason). Assertion: The cube of 15 has more digits than the square of 15. Reason: For any number greater than 1, its cube is always numerically larger than its square.
   Solution: Both statements are true and the reason correctly explains the assertion: 15²=225 (3 digits), 15³=3375 (4 digits), and since 15>1, 15³ > 15².
6. 6 (Numerical). Without adding term by term, find the sum: 1+3+5+7+9+11+13+15+17.
   Solution: This is the sum of the first 9 consecutive odd numbers, which equals 9² = 81.
7. 7 (Synthesis). Is 2025 both a perfect square and a perfect cube? Justify using prime factorisation.
   Solution: 2025 = 3⁴×5². Both exponents (4 and 2) are even, so 2025 is a perfect square (45²). But for a perfect cube every exponent must be a multiple of 3 — 4 and 2 are not — so 2025 is not a perfect cube.
8. 8 (Applied). How many numbers lie between 45² and 46²?
   Solution: Using the rule that 2n numbers lie between n² and (n+1)²: 2×45 = 90 numbers.
9. 9 (Assertion-Reason). Assertion: 216 is a perfect cube. Reason: 216 = 2³×3³.
   Solution: Both true, and the reason correctly explains the assertion: since both prime factors appear in triples, 216 = 6³ is indeed a perfect cube.
10. 10 (Numerical). Find the smallest 4-digit perfect square.
    Solution: √1000 ≈ 31.6, so the next whole number is 32. 32² = 1024, the smallest 4-digit perfect square.

Written by Satish

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