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.

Last Updated: September 23, 2026

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.

Frequently Asked Questions

How can you find the square root of a perfect square using the prime factorisation method?
The number is broken down into its prime factors, and since a perfect square has each prime factor appearing an even number of times, one factor from each pair is taken and multiplied together to get the square root.

Why is the cube root of a negative number also negative, unlike the square root?
Multiplying three negative numbers together gives a negative result, so the cube root of a negative number is defined and negative, whereas the square root of a negative number has no real value.

Written by Satish

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