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SparkEd Math offers 60+ practice questions for Squares, Cubes & Their Roots across 3 difficulty levels (Easy, Medium, Hard) plus a timed test mode. Each question comes with visual step-by-step solutions and AI coaching hints.

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SparkEd Math uses visual step-by-step solutions for every Squares, Cubes & Their Roots problem, making complex concepts easy to understand. The platform includes 3 difficulty levels for progressive learning, a Super Power Help AI coach for instant doubt-clearing, and interactive practice rather than passive reading.

Is SparkEd's Squares, Cubes & Their Roots content aligned with the CBSE syllabus?

Yes, all Squares, Cubes & Their Roots questions and lessons on SparkEd Math are fully aligned with the Central Board of Secondary Education (CBSE) syllabus for Class 8. The content follows the official curriculum structure and exam pattern.

Squares, Cubes & Their Roots — Class 8 CBSE (2025-26)

About Squares, Cubes & Their Roots — Class 8 CBSE

Find squares, cubes, square roots, and cube roots using various methods. This topic is part of the CBSE Class 8 mathematics syllabus (chapter: Chapter 1). On this page you can practice 60 questions across three difficulty levels — 20 easy, 20 medium, and 20 hard — each with a visual step-by-step solution, plus a timed 35-question mastery test with a live leaderboard. Worked examples and sample questions from the topic are below.

Squares, Cubes & Their Roots — solved examples for Class 8 CBSE

Example 1easy

Which of the following numbers, based on its unit digit, can definitively be stated as NOT a perfect square?

A)3136
B)4096
C)5249
D)6723

Step-by-step solution

A perfect square cannot end with the digits 2, 3, 7, or 8.
Let's check the unit digit of each option:
3136 ends in 6 (can be a perfect square, e.g., 56²).
4096 ends in 6 (can be a perfect square, e.g., 64²).
5249 ends in 9 (can be a perfect square, e.g., 73²).
6723 ends in 3. Since a perfect square cannot end in 3, 6723 cannot be a perfect square.

Answer: 6723

Example 2medium

Which of the following numbers cannot be a perfect square?

A)2401
B)3969
C)5776
D)7208

Step-by-step solution

A perfect square can only end with the digits 0, 1, 4, 5, 6, or 9.
The number 7208 ends with the digit 8.
Therefore, 7208 cannot be a perfect square.
The other numbers 2401 (ends in 1), 3969 (ends in 9), and 5776 (ends in 6) can be perfect squares. (Specifically, 49², 63², and 76² respectively).

Answer: 7208

Example 3hard

The sum of the first 'n' odd natural numbers is 2025. What is the value of 'n'?

A)A) 35
B)B) 40
C)C) 45
D)D) 50

Step-by-step solution

The sum of the first 'n' odd natural numbers is given by the formula n².
Given that the sum is 2025, we have the equation n² = 2025.
To find 'n', we need to calculate the square root of 2025.
$n = √2025$
By prime factorization or estimation (since 40² = 1600 and 50² = 2500, and 2025 ends in 5, 'n' must end in 5), we find that 45 × 45 = 2025. Thus, n = 45.

Answer: C) 45

Practice questions on Squares, Cubes & Their Roots

Q1.easy
Which statement is TRUE about the number of non-perfect square numbers between n² and (n+1)² for any natural number 'n'?
1. A)It is always 2n.
2. B)It is always n+1.
3. C)It is always 2n-1.
4. D)It depends on the specific value of n.
Show answer
Answer: It is always 2n.
Hint: Consider a small example, like the numbers between 2² and 3².
Q2.easy
Which of the following represents the sum of the first 6 odd natural numbers?
1. A)36
2. B)25
3. C)49
4. D)64
Show answer
Answer: 36
Hint: Remember the pattern relating the sum of consecutive odd numbers to square numbers.
Q3.easy
Ravi tried to find the square root of 196 by prime factorization. His steps are shown below:
1. 196 = 2 × 98
2. = 2 × 2 × 49
3. = 2 × 2 × 7 × 7
4. He then wrote, √196 = 2 × 7 = 14.
Which of the following statements about Ravi's solution is correct?
1. A)Ravi correctly applied the prime factorization method to find the square root.
2. B)Ravi made a mistake in the prime factorization in step 3.
3. C)Ravi should have paired the factors as (2×7) and (2×7) before taking one from each pair.
4. D)The square root of 196 is not 14.
Show answer
Answer: Ravi correctly applied the prime factorization method to find the square root.
Hint: Review the steps involved in finding the square root using prime factorization.
Q4.medium
Evaluate (105)² using a suitable algebraic identity or pattern.
1. A)10250
2. B)11025
3. C)10525
4. D)11500
Show answer
Answer: 11025
Hint: Consider expressing 105 as a sum, for example, (100 + 5), and then apply a known algebraic identity.
Q5.medium
A general wishes to arrange his 10,500 soldiers in the form of a solid square. After arrangement, he finds some soldiers are left out. Find how many soldiers are left out.
1. A)24 soldiers
2. B)50 soldiers
3. C)76 soldiers
4. D)100 soldiers
Show answer
Answer: 76 soldiers
Hint: To form a solid square, the number of soldiers must be a perfect square. Find the largest perfect square less than 10,500.
Q6.medium
Find the smallest whole number by which 252 must be multiplied to get a perfect square. What is the square root of the new number?
1. A)7; √1764 = 42
2. B)7; √252 = 15.87
3. C)3; √756 = 27.5
4. D)6; √1512 = 38.8
Show answer
Answer: 7; √1764 = 42
Hint: Use the prime factorization method. For a number to be a perfect square, all its prime factors must appear in pairs.
Q7.hard
Which of the following statements is ALWAYS TRUE for a perfect square 'N'?
1. A)A) N must end with an even number of zeroes.
2. B)B) N cannot have 2, 3, 7, or 8 as its unit digit.
3. C)C) If N is divisible by 3, then it must be divisible by 9.
4. D)D) All of the above.
Show answer
Answer: D) All of the above.
Hint: Carefully evaluate each statement for all possible perfect squares. Look for any counter-examples before making a conclusion.
Q8.hard
If (2m-1) is one of the members of a Pythagorean triplet, and it is the smallest odd member, which of the following could be the largest member of that triplet?
1. A)A) 2m² - 2m + 1
2. B)B) 2m² + 2m + 1
3. C)C) 4m² - 4m + 1
4. D)D) 4m² - 4m + 2
Show answer
Answer: A) 2m² - 2m + 1
Hint: Recall the general form of a Pythagorean triplet when one member is an odd number 'x'. The other two members are (x²-1)/2 and (x²+1)/2.
Q9.hard
What is the least number that must be subtracted from 13860 to make it a perfect square?
1. A)A) 100
2. B)B) 121
3. C)C) 171
4. D)D) 189
Show answer
Answer: C) 171
Hint: Use the long division method to find the square root of 13860. The remainder will be the number to subtract.

These are 9 of the 60 questions available for Squares, Cubes & Their Roots. Start practicing above to unlock visual solutions, AI coaching, and the topic leaderboard — completely free.