Sample Paper 2 Class8 A Square And A Cube
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CLASS VIII • MATHEMATICS PRACTICE TEST (2026–27)
Topic: Chapter 1 — A Square and A Cube (Ganita Prakash)
- All questions are compulsory.
- This question paper contains 20 questions divided into five Sections: A, B, C, D, and E.
- Section A contains 10 Multiple Choice Questions of 1 mark each (Q1 to Q10, including 2 Assertion-Reason questions).
- Section B contains 4 Short Answer questions of 2 marks each (Q11 to Q14).
- Section C contains 3 Short Answer questions of 3 marks each (Q15 to Q17).
- Section D contains 1 Long Answer question of 5 marks (Q18).
- Section E contains 2 Case Study-based questions of 4 marks each (Q19 & Q20).
- Use of calculators is strictly prohibited.
1. Which of the following numbers can definitely NOT be a perfect square?
[1]
✓ Solution (1 Mark):
A number ending with digits 2, 3, 7, or 8 is never a perfect square. The number 7892 ends in 2, so it cannot be a perfect square. (Correct Option: c)
2. In a locker room with 144 lockers numbered 1 to 144, each person toggles lockers matching multiples of their turn. How many lockers remain open at the end?
[1]
✓ Solution (1 Mark):
Only square numbers have an odd number of factors and remain open. Since 144 = 12², the lockers open are 1², 2², 3², ..., 12² (Total = 12 lockers). (Correct Option: b)
3. If a natural number ends with 3 zeros, how many zeros will its cube have at the end?
[1]
✓ Solution (1 Mark):
If a number has n zeros at the end, its cube contains 3n zeros. For n = 3, the number of zeros = 3 × 3 = 9. (Correct Option: c)
4. How many non-square natural numbers lie strictly between 15² and 16²?
[1]
✓ Solution (1 Mark):
The number of non-square numbers between n² and (n + 1)² is given by 2n. Here n = 15, so 2 × 15 = 30. (Correct Option: a)
5. The sum of two consecutive triangular numbers 10 and 15 is:
[1]
✓ Solution (1 Mark):
The sum of any two consecutive triangular numbers always produces a perfect square: 10 + 15 = 25 (which is 5²). (Correct Option: b)
6. What is the units digit of the cube of a number ending in 7?
[1]
✓ Solution (1 Mark):
7 × 7 × 7 = 343. Thus, any integer ending in 7 produces a cube that ends in the digit 3. (Correct Option: b)
7. The sum 31 + 33 + 35 + 37 + 39 + 41 without actual addition is equal to:
[1]
✓ Solution (1 Mark):
This is the sum of 6 consecutive odd numbers starting from [6 × 5 + 1] = 31. By the cube-odd pattern, this sum equals 6³ = 216. (Correct Option: c)
8. The smallest natural number by which 675 must be multiplied to make it a perfect cube is:
[1]
✓ Solution (1 Mark):
Prime factorisation of 675 = 3 × 3 × 3 × 5 × 5 = 3³ × 5². To complete the triplet of 5, we must multiply by 5. (Correct Option: b)
9. Assertion (A): The square of 0.7 is 0.49, which is smaller than 0.7.
Reason (R): The square of any proper fraction or decimal between 0 and 1 is always less than the original number. [1]
Reason (R): The square of any proper fraction or decimal between 0 and 1 is always less than the original number. [1]
✓ Solution (1 Mark):
(0.7)² = 0.49 < 0.7. For any x where 0 < x < 1, multiplying x by itself yields x² < x. Both statements are true and R is the direct explanation of A. (Correct Option: a)
10. Assertion (A): The number 2400 is not a perfect cube.
Reason (R): A natural number ending with an even number of zeros can never be a perfect square. [1]
Reason (R): A natural number ending with an even number of zeros can never be a perfect square. [1]
✓ Solution (1 Mark):
Assertion A is true because 2400 ends in two zeros (a cube cannot end in 2 zeros). Reason R is false because perfect squares CAN end in an even number of zeros (e.g., 100 = 10², 400 = 20²). (Correct Option: c)
11. By what least number should 1800 be divided to get a perfect square? Find the square root of the quotient.
[2]
✓ Solution (2 Marks):
Resolving 1800 into prime factors:
1800 = 2 × 2 × 2 × 3 × 3 × 5 × 5 = (2²) × (3²) × (5²) × 2.
The prime factor 2 is unpaired. Therefore, 1800 must be divided by 2.
New quotient = 1800 ÷ 2 = 900.
Square root = √900 = 2 × 3 × 5 = 30.
1800 = 2 × 2 × 2 × 3 × 3 × 5 × 5 = (2²) × (3²) × (5²) × 2.
The prime factor 2 is unpaired. Therefore, 1800 must be divided by 2.
New quotient = 1800 ÷ 2 = 900.
Square root = √900 = 2 × 3 × 5 = 30.
12. Find the cube root of 5832 using the prime factorisation method.
[2]
✓ Solution (2 Marks):
Prime factorisation of 5832:
5832 = (2 × 2 × 2) × (3 × 3 × 3) × (3 × 3 × 3) = 2³ × 3³ × 3³.
³√5832 = 2 × 3 × 3 = 18.
5832 = (2 × 2 × 2) × (3 × 3 × 3) × (3 × 3 × 3) = 2³ × 3³ × 3³.
³√5832 = 2 × 3 × 3 = 18.
13. Given that 65² = 4225, calculate the value of 66² using consecutive square identities rather than column multiplication.
[2]
✓ Solution (2 Marks):
Using the pattern: (n + 1)² = n² + 2n + 1 (or n² + n + (n + 1)):
66² = 65² + (65 + 66) = 4225 + 131 = 4356.
Alternatively: 66² = 4225 + 2(65) + 1 = 4225 + 130 + 1 = 4356.
66² = 65² + (65 + 66) = 4225 + 131 = 4356.
Alternatively: 66² = 4225 + 2(65) + 1 = 4225 + 130 + 1 = 4356.
14. In an assembly hall, 784 chairs are arranged in a square grid such that the number of rows equals the number of chairs in each row. How many rows are there?
[2]
✓ Solution (2 Marks):
Let the number of rows be x. Number of chairs in each row = x.
Total chairs = x × x = x² = 784.
x = √784 = √(2 × 2 × 2 × 2 × 7 × 7) = 2 × 2 × 7 = 28 rows.
Total chairs = x × x = x² = 784.
x = √784 = √(2 × 2 × 2 × 2 × 7 × 7) = 2 × 2 × 7 = 28 rows.
15. Find the smallest square number that is completely divisible by each of the numbers 8, 12, and 20.
[3]
✓ Solution (3 Marks):
Step 1: Find the LCM of 8, 12, and 20.
8 = 2³, 12 = 2² × 3, 20 = 2² × 5.
LCM = 2³ × 3 × 5 = 120.
Step 2: Express LCM as prime factors: 120 = (2²) × 2 × 3 × 5.
To make it a perfect square, each factor must be in pairs. We must multiply by 2 × 3 × 5 = 30.
Required square number = 120 × 30 = 3600.
8 = 2³, 12 = 2² × 3, 20 = 2² × 5.
LCM = 2³ × 3 × 5 = 120.
Step 2: Express LCM as prime factors: 120 = (2²) × 2 × 3 × 5.
To make it a perfect square, each factor must be in pairs. We must multiply by 2 × 3 × 5 = 30.
Required square number = 120 × 30 = 3600.
16. Find the smallest natural number by which 3087 must be multiplied so that the product becomes a perfect cube. Also find the cube root of the product.
[3]
✓ Solution (3 Marks):
Prime factorisation of 3087: 3087 = 3 × 3 × 7 × 7 × 7 = 3² × 7³.
The factor 3 occurs only twice. To complete the triplet, 3087 must be multiplied by 3.
New product = 3087 × 3 = 9261.
Cube root = ³√9261 = 3 × 7 = 21.
The factor 3 occurs only twice. To complete the triplet, 3087 must be multiplied by 3.
New product = 3087 × 3 = 9261.
Cube root = ³√9261 = 3 × 7 = 21.
17. In the textbook Ganita Prakash, the concept of successive differences is explored. For the first five cubes (1, 8, 27, 64, 125), calculate the Level 1, Level 2, and Level 3 differences, and state what is observed at Level 3.
[3]
✓ Solution (3 Marks):
Cubes: 1, 8, 27, 64, 125.
Level 1 Differences: (8 - 1) = 7; (27 - 8) = 19; (64 - 27) = 37; (125 - 64) = 61.
Level 2 Differences: (19 - 7) = 12; (37 - 19) = 18; (61 - 37) = 24.
Level 3 Differences: (18 - 12) = 6; (24 - 18) = 6.
Observation: At Level 3, all successive differences are constant and equal to 6.
Level 1 Differences: (8 - 1) = 7; (27 - 8) = 19; (64 - 27) = 37; (125 - 64) = 61.
Level 2 Differences: (19 - 7) = 12; (37 - 19) = 18; (61 - 37) = 24.
Level 3 Differences: (18 - 12) = 6; (24 - 18) = 6.
Observation: At Level 3, all successive differences are constant and equal to 6.
18. (a) Evaluate √5476 using the long division method. [2 Marks]
(b) A school gardener has 2800 flowering saplings. He wants to plant them in a square grid with an equal number of rows and columns. What is the minimum number of saplings that must be added to achieve this arrangement? Find the total number of saplings and the number of plants in each row. [3 Marks] [5]
(b) A school gardener has 2800 flowering saplings. He wants to plant them in a square grid with an equal number of rows and columns. What is the minimum number of saplings that must be added to achieve this arrangement? Find the total number of saplings and the number of plants in each row. [3 Marks] [5]
✓ Solution (5 Marks):
(a) Long division for √5476:
Group in pairs: 54 76.
7 × 7 = 49. Remainder = 54 - 49 = 5. Bring down 76 ⇒ 576.
Double 7 = 14. Test digit 4: 144 × 4 = 576. Remainder = 0.
Thus, √5476 = 74. [2 Marks]
(b) Minimum saplings to add to 2800:
Testing squares: 52² = 2704, and 53² = 2809.
Since 2704 < 2800 < 2809, the next perfect square is 53² = 2809.
Minimum saplings to be added = 2809 - 2800 = 9 saplings.
Total saplings after addition = 2809.
Number of plants in each row = 53. [3 Marks]
Group in pairs: 54 76.
7 × 7 = 49. Remainder = 54 - 49 = 5. Bring down 76 ⇒ 576.
Double 7 = 14. Test digit 4: 144 × 4 = 576. Remainder = 0.
Thus, √5476 = 74. [2 Marks]
(b) Minimum saplings to add to 2800:
Testing squares: 52² = 2704, and 53² = 2809.
Since 2704 < 2800 < 2809, the next perfect square is 53² = 2809.
Minimum saplings to be added = 2809 - 2800 = 9 saplings.
Total saplings after addition = 2809.
Number of plants in each row = 53. [3 Marks]
19. Case Study 1: The School Botanical Square Grid Project
[4]
Students of the eco-club in Class 8 are designing square botanical plots. They learned that square numbers can be verified through successive subtraction of consecutive odd numbers, and that their boundaries follow exact integral relationships known as Pythagorean triplets.
(a) Show whether 64 is a perfect square by using the method of successive subtraction of consecutive odd numbers starting from 1. [1 Mark]
(b) Write a Pythagorean triplet whose smallest member is 10. [1 Mark]
(c) An ornamental display floor in the garden is paved with 1521 square ceramic tiles. Find the number of tiles along one full side of the square floor. [1 Mark]
(d) How many non-square numbers lie strictly between 24² and 25²? [1 Mark]
✓ Solution (4 Marks):
(a) Successive subtractions: 64 - 1 = 63; 63 - 3 = 60; 60 - 5 = 55; 55 - 7 = 48; 48 - 9 = 39; 39 - 11 = 28; 28 - 13 = 15; 15 - 15 = 0. Since 0 is reached at the 8th step, 64 is a perfect square and √64 = 8. [1 Mark]
(b) For smallest member 2m = 10 ⇒ m = 5. Other members: m² - 1 = 25 - 1 = 24, and m² + 1 = 25 + 1 = 26. Triplet: (10, 24, 26). [1 Mark]
(c) Side = √1521. By prime factorisation, 1521 = 3² × 13². Side = 3 × 13 = 39 tiles. [1 Mark]
(d) Non-square numbers = 2n = 2 × 24 = 48. [1 Mark]
(b) For smallest member 2m = 10 ⇒ m = 5. Other members: m² - 1 = 25 - 1 = 24, and m² + 1 = 25 + 1 = 26. Triplet: (10, 24, 26). [1 Mark]
(c) Side = √1521. By prime factorisation, 1521 = 3² × 13². Side = 3 × 13 = 39 tiles. [1 Mark]
(d) Non-square numbers = 2n = 2 × 24 = 48. [1 Mark]
20. Case Study 2: The Enigma of Taxicab Numbers and Cube Properties
[4]
During a mathematics colloquium, students studied the famous dialogue between G.H. Hardy and Srinivasa Ramanujan regarding the taxi number 1729. Ramanujan explained that 1729 is the smallest number that can be expressed as the sum of two positive cubes in two different ways (1³ + 12³ = 9³ + 10³). Such numbers are called Hardy-Ramanujan or Taxicab numbers.
(a) The third taxicab number is 13832. Express 13832 as the sum of two cubes in two different ways using numbers from {2, 18, 20, 24}. [1 Mark]
(b) Without actual addition, determine the sum: 57 + 59 + 61 + 63 + 65 + 67 + 69 + 71. [1 Mark]
(c) Find the smallest natural number by which 2560 must be multiplied so that the product becomes a perfect cube. [1 Mark]
(d) Which is greater: (51³ - 50³) or (51² - 50²)? Justify your answer without computing large powers. [1 Mark]
✓ Solution (4 Marks):
(a) 13832 can be written as: 2³ + 24³ (8 + 13824 = 13832) and 18³ + 20³ (5832 + 8000 = 13832). [1 Mark]
(b) There are 8 consecutive odd numbers starting from (8 × 7 + 1) = 57. By the consecutive odd cube identity, this sum is equal to 8³ = 512. [1 Mark]
(c) 2560 = 256 × 10 = 2⁸ × 2 × 5 = 2⁹ × 5¹. The prime factor 5 needs two more 5s (5²) to form a triplet. Smallest multiplier = 5² = 25. [1 Mark]
(d) (51³ - 50³) = (51 - 50)(51² + 50² + 51 × 50) = 51² + 50² + 2550 > 5000.
(51² - 50²) = (51 - 50)(51 + 50) = 101.
Clearly, (51³ - 50³) is greater. [1 Mark]
(b) There are 8 consecutive odd numbers starting from (8 × 7 + 1) = 57. By the consecutive odd cube identity, this sum is equal to 8³ = 512. [1 Mark]
(c) 2560 = 256 × 10 = 2⁸ × 2 × 5 = 2⁹ × 5¹. The prime factor 5 needs two more 5s (5²) to form a triplet. Smallest multiplier = 5² = 25. [1 Mark]
(d) (51³ - 50³) = (51 - 50)(51² + 50² + 51 × 50) = 51² + 50² + 2550 > 5000.
(51² - 50²) = (51 - 50)(51 + 50) = 101.
Clearly, (51³ - 50³) is greater. [1 Mark]