Chapter-wise Worksheets for Class 8 Mathematics: Ganita Prakash Part 1 Chapter 01 A Square and A Cube
Explore structured practice materials through the CBSE Class 8 Mathematics A Square and A Cube Practice Paper Worksheet Set 02 with Answers. Tailored for Class 8 learners, utilizing these Mathematics worksheets ensures thorough preparation and strengthens problem-solving accuracy before final school evaluations.
Practice Class 8 Mathematics Worksheets: Ganita Prakash Part 1 Chapter 01 A Square and A Cube
View or download the dedicated CBSE Class 8 Mathematics A Square and A Cube Practice Paper Worksheet Set 02 with Answers resource below. Engaging with these practice papers under focused study conditions ensures continuous academic progress and mastery of the 2026-27 curriculum for Ganita Prakash Part 1 Chapter 01 A Square and A Cube.
Practice Paper 07 on CBSE Class 8 Maths Ganita Prakash Chapter 1 A Square and A Cube carries 40 marks over 20 questions in Sections A to E, and every question is solved step by step below. It tests units digits and end zeros of squares, the count of non-squares between consecutive squares, the sum of the first n odd numbers, why the number of factors of a cube can be odd or even, square and cube roots by prime factorisation, and word problems on a class fund, a square formation of soldiers and building the nearest perfect cube from 5400 blocks.
Section A: Multiple Choice Questions (1 Mark Each)
(a) It is a perfect square.
(b) It is not a perfect square.
(c) It can be a perfect square, but only if the other digits form a perfect square.
(d) Cannot be determined.
Show Answer & Explanation
Answer: (b) It is not a perfect square.
Explanation:
1. The units digit of a square is the units digit of 0², 1², …, 9², that is 0, 1, 4, 9, 6, 5, 6, 9, 4, 1.
2. So a square can only end in 0, 1, 4, 5, 6 or 9, never in 2, 3, 7 or 8.
3. A number ending in 2 is not a perfect square.
Teacher's Note:
1. Numbers ending in 2, 3, 7 or 8 can be rejected as squares at a glance.
2. The opposite is not true: ending in 1, 4, 5, 6, 9 or 0 does not guarantee a square.
(a) The number of zeros must be odd.
(b) The number of zeros must be even.
(c) The number of zeros must be 0.
(d) The number of zeros can be either even or odd.
Show Answer & Explanation
Answer: (b) The number of zeros must be even.
Explanation:
1. End zeros come from factors of 10, and in a square every factor appears in pairs.
2. So the zeros at the end of a square come in pairs: 10² = 100 (2 zeros), 100² = 10000 (4 zeros).
3. The number of zeros must be even (0 zeros, as in 49, also counts as even).
Teacher's Note:
1. A number ending in 1 or 3 zeros (like 10 or 1000) can never be a square.
2. Do not confuse with cubes, where end zeros come in groups of three.
(a) 2.25 square units
(b) 3.0 square units
(c) 2.5 square units
(d) 1.5 square units
Show Answer & Explanation
Answer: (a) 2.25 square units
Explanation:
1. Area of a square = side × side.
2. 1.5 × 1.5 = 2.25.
3. So the area is 2.25 square units.
Teacher's Note:
1. 3.0 is 2 × 1.5, the mistake of doubling instead of squaring.
2. 15 × 15 = 225, and there are two decimal places in total, so 1.5² = 2.25.
(a) 14
(b) 28
(c) 29
(d) 30
Show Answer & Explanation
Answer: (b) 28
Explanation:
1. 14² = 196 and 15² = 225.
2. The numbers strictly between them are 197, 198, …, 224, which is 224 − 197 + 1 = 28 numbers.
3. In general there are 2n non-squares between n² and (n + 1)²; here 2 × 14 = 28.
Teacher's Note:
1. The gap between n² and (n + 1)² is 2n + 1, and the numbers strictly in between are one less: 2n.
2. None of the numbers between two consecutive squares is a square.
(a) 8
(b) 6
(c) 4
(d) 2
Show Answer & Explanation
Answer: (d) 2
Explanation:
1. 8 × 8 = 64, which ends in 4.
2. 4 × 8 = 32, which ends in 2.
3. So 8³ ends in 2 (8³ = 512).
Teacher's Note:
1. Cubes of numbers ending in 2 end in 8, and cubes of numbers ending in 8 end in 2.
2. Track only the units digit at each step of the multiplication.
(a) It must have an even number of factors.
(b) It must have an odd number of factors.
(c) It can have either an odd or even number of factors.
(d) It must have a prime number of factors.
Show Answer & Explanation
Answer: (c) It can have either an odd or even number of factors.
Explanation:
1. 8 = 2³ has factors 1, 2, 4, 8, which is 4 factors (even).
2. 64 = 4³ has factors 1, 2, 4, 8, 16, 32, 64, which is 7 factors (odd), because 64 is also a square (8²).
3. So a perfect cube can have either an odd or an even number of factors.
4. 8 has 4 factors, which is not prime, so (d) is wrong too.
Teacher's Note:
1. The source key gave (a), but cubes that are also squares, like 1, 64 and 729, have an odd number of factors, so the key is corrected to (c).
2. The odd-or-even rule for factors depends only on whether a number is a square.
(a) 7 units
(b) 8 units
(c) 9 units
(d) 10 units
Show Answer & Explanation
Answer: (b) 8 units
Explanation:
1. Volume of a cube = side × side × side, so side = ∛volume.
2. 512 = 2⁹ = (2 × 2 × 2)³ = 8³.
3. So the side length is 8 units.
Teacher's Note:
1. Check: 8 × 8 × 8 = 64 × 8 = 512.
2. 7³ = 343 and 9³ = 729, so 512 must be 8³.
(a) 10
(b) 12
(c) 14
(d) 16
Show Answer & Explanation
Answer: (a) 10
Explanation:
1. The sum of the first n odd numbers is n².
2. n² = 100, and 10 × 10 = 100.
3. So n = 10; indeed 1 + 3 + 5 + … + 19 = 100.
Teacher's Note:
1. The last of the first n odd numbers is 2n − 1, which is 19 here.
2. Think of 100 dots arranged as a 10 × 10 square built from L-shaped layers.
Assertion (A): The sum of the first 6 odd natural numbers is 36.
Reason (R): The sum of the first n odd natural numbers is n².
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. 1 + 3 + 5 + 7 + 9 + 11 = 36, so A is true.
2. R is the known pattern: 1 = 1², 1 + 3 = 2², 1 + 3 + 5 = 3², and so on. R is true.
3. Putting n = 6 in R gives 6² = 36, which is exactly A. Answer: (a).
Teacher's Note:
1. A is a special case of R, so R explains A.
2. Each new odd number adds an L-shaped layer that turns an n × n square into an (n + 1) × (n + 1) square.
Assertion (A): The number 729 is a perfect cube.
Reason (R): The prime factorization of 729 is 3⁶, and its prime factors can be grouped into three identical sets.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. 729 = 3 × 3 × 3 × 3 × 3 × 3 = 3⁶, so the first part of R is true.
2. Group the six 3s into three identical sets: (3 × 3) × (3 × 3) × (3 × 3) = 9 × 9 × 9.
3. So 729 = 9³ is a perfect cube; A is true.
4. Splitting into three identical sets is what makes a number a cube, so R explains A. Answer: (a).
Teacher's Note:
1. 729 is both a cube (9³) and a square (27²).
2. Six equal prime factors can be split into three groups of two or two groups of three.
Section B: Short Answer Questions (2 Marks Each)
Question 11. Find the smallest square number which is a multiple of 8, 12, and 18.
Answer:
8 = 2 × 2 × 2, 12 = 2 × 2 × 3, 18 = 2 × 3 × 3, so LCM = 2 × 2 × 2 × 3 × 3 = 72.
72 = (2 × 2) × 2 × (3 × 3); one factor 2 is unpaired.
Multiply by 2: 72 × 2 = 144 = 12².
The smallest square number is 144.
In simple words: LCM of 8, 12 and 18 is 72; one more 2 makes it the square 144.
Teacher's Note:
1. Check: 144 ÷ 8 = 18, 144 ÷ 12 = 12, 144 ÷ 18 = 8.
2. Multiply only by what is needed to pair up every factor.
Question 12. Find the square root of 576 by the prime factorisation method.
Answer:
576 ÷ 2 = 288, ÷ 2 = 144, ÷ 2 = 72, ÷ 2 = 36, ÷ 2 = 18, ÷ 2 = 9, ÷ 3 = 3, ÷ 3 = 1.
576 = (2 × 2) × (2 × 2) × (2 × 2) × (3 × 3).
√576 = 2 × 2 × 2 × 3 = 24.
In simple words: Pair up the six 2s and two 3s and take one from each pair: 24.
Teacher's Note:
1. Check: 24 × 24 = 576.
2. 576 ends in 6, so its square root ends in 4 or 6; 24 fits.
Question 13. Find the cube root of 5832 by the prime factorization method.
Answer:
5832 ÷ 2 = 2916, ÷ 2 = 1458, ÷ 2 = 729, and 729 = 3 × 3 × 3 × 3 × 3 × 3.
5832 = (2 × 2 × 2) × (3 × 3 × 3) × (3 × 3 × 3).
∛5832 = 2 × 3 × 3 = 18.
In simple words: Group the prime factors of 5832 in threes and take one from each group: 18.
Teacher's Note:
1. Check: 18 × 18 × 18 = 324 × 18 = 5832.
2. 5832 ends in 2, so its cube root ends in 8; this agrees with 18.
Question 14. What is the smallest number by which 121 must be multiplied so that the product is a perfect cube?
Answer:
121 = 11 × 11, only two 11s.
One more 11 completes the triplet: 121 × 11 = 1331 = 11³.
The smallest multiplier is 11.
In simple words: 121 is 11 × 11; multiply by one more 11 to get the cube 1331.
Teacher's Note:
1. 121 is a square (11²) but needs one more 11 to become a cube.
2. 11³ = 1331 is a cube worth remembering.
Section C: Short Answer Questions (3 Marks Each)
Question 15. Find the smallest number by which 288 must be divided to obtain a perfect square. Also, find the square root of the quotient.
Answer:
288 = 2 × 2 × 2 × 2 × 2 × 3 × 3 = (2 × 2) × (2 × 2) × 2 × (3 × 3).
One factor 2 is unpaired, so divide by 2: 288 ÷ 2 = 144.
144 = (2 × 2) × (2 × 2) × (3 × 3), so √144 = 2 × 2 × 3 = 12.
The smallest divisor is 2 and the square root of the quotient is 12.
In simple words: Divide out the extra 2 to get 144, whose square root is 12.
Teacher's Note:
1. Five 2s means one 2 is left after making pairs.
2. Check: 12 × 12 = 144 and 144 × 2 = 288.
Question 16. Find the smallest number by which 3267 must be divided so that the quotient has a cube root. Also, find the cube root of the quotient.
Answer:
3267 ÷ 3 = 1089, ÷ 3 = 363, ÷ 3 = 121, ÷ 11 = 11, ÷ 11 = 1.
3267 = (3 × 3 × 3) × 11 × 11. The 3s form a triplet; the two 11s do not.
Divide by 11 × 11 = 121: 3267 ÷ 121 = 27.
27 = 3 × 3 × 3, so ∛27 = 3.
The smallest divisor is 121 and the cube root of the quotient is 3.
In simple words: Remove both 11s by dividing by 121; 27 = 3³ is left.
Teacher's Note:
1. Dividing by just 11 leaves one 11, which is still not a triplet.
2. The digit sum of 3267 is 18, so it is divisible by 9; that helps start the factorisation.
Question 17. Find the smallest number which when multiplied with 540 will make the product a perfect cube. Find the cube root of the product.
Answer:
540 = 2 × 2 × 3 × 3 × 3 × 5.
The 3s form a triplet. The 2s need one more 2, and the 5 needs two more 5s.
Multiply by 2 × 5 × 5 = 50: 540 × 50 = 27000.
27000 = (2 × 2 × 2) × (3 × 3 × 3) × (5 × 5 × 5), so ∛27000 = 2 × 3 × 5 = 30.
The smallest multiplier is 50 and the cube root of the product is 30.
In simple words: Add one 2 and two 5s (multiply by 50) to get 27000 = 30³.
Teacher's Note:
1. 27000 = 27 × 1000 = 3³ × 10³ = 30³, a quick check.
2. Each prime needs its own number of extra copies to reach a multiple of three.
Section D: Long Answer Question (5 Marks)
Question 18. The students of a class collected 2601 rupees for a fund. Each student contributed as many rupees as there were students in the class. Find the number of students in the class and the amount each student contributed.
Answer:
Let the number of students be x. Each student gave x rupees.
Total collected = number of students × amount per student = x × x = x².
So x² = 2601 and x = √2601.
Prime factorisation: 2601 ÷ 3 = 867, ÷ 3 = 289, ÷ 17 = 17, ÷ 17 = 1.
2601 = (3 × 3) × (17 × 17).
√2601 = 3 × 17 = 51.
There are 51 students and each student contributed ₹51.
In simple words: Students × rupees each = 2601, and both are the same number, so it is √2601 = 51.
Teacher's Note:
1. "As many rupees as there were students" means the total is a perfect square.
2. Check: 51 × 51 = 2601.
3. The digit sum 2 + 6 + 0 + 1 = 9 shows that 2601 is divisible by 9, a good starting point.
Section E: Case Study Based Questions (4 Marks Each)
Question 19. Case Study 1 (The King's Fort): King Vikramaditya wanted to build a new fort in the shape of a perfect square. He had 1500 soldiers and wanted to arrange them in a solid square formation, with the same number of rows as soldiers in each row. He realized that with his current number of soldiers, he would be left with a few extra soldiers. He wanted to know how many more soldiers he would need to form a perfect square with no extra soldiers.
(i) What is the largest perfect square number of soldiers less than 1500?
(ii) How many soldiers would be left over?
(iii) What is the smallest perfect square number of soldiers greater than 1500?
(iv) How many more soldiers are needed to form the next perfect square?
Answer:
Use known squares: 30² = 900 and 40² = 1600, so √1500 lies between 30 and 40, close to 40.
38² = 1444 and 39² = 1521.
(i) The largest perfect square less than 1500 is 1444 (38 rows of 38 soldiers).
(ii) Soldiers left over = 1500 − 1444 = 56.
(iii) The smallest perfect square greater than 1500 is 1521 (39 × 39).
(iv) More soldiers needed = 1521 − 1500 = 21.
In simple words: 1500 lies between 38² = 1444 and 39² = 1521, so 56 are extra or 21 more are needed.
Teacher's Note:
1. The original passage said the soldiers stand on the four sides of the fort; that would only need 1500 ÷ 4 = 375 per side and leaves no extra soldiers, so it is reworded as a solid square formation to match the questions.
2. Check: 56 + 21 = 77 = 2 × 38 + 1, the gap between 38² and 39².
Question 20. Case Study 2 (The Building Block Challenge): A young builder named Maya is given a large number of cubical wooden blocks. She is challenged to arrange them to form a large perfect cube. She has a total of 5400 blocks. She realizes she needs to either add or remove some blocks to form a perfect cube.
(i) What is the prime factorization of 5400?
(ii) What is the least number of blocks that must be removed to make it a perfect cube?
(iii) What will be the side length of the new perfect cube formed by adding the minimum number of blocks?
Answer:
(i) 5400 = 54 × 100 = (2 × 3 × 3 × 3) × (2 × 2 × 5 × 5).
So 5400 = 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5 = 2³ × 3³ × 5². The two 5s are not a triplet, so 5400 is not a cube.
(ii) Find the cubes on either side of 5400: 17³ = 4913 and 18³ = 5832.
Removing blocks gives the largest cube below 5400, which is 4913.
Least number of blocks to remove = 5400 − 4913 = 487 (a cube of side 17).
(iii) Adding blocks gives the smallest cube above 5400, which is 5832 = 18³.
Blocks to add = 5832 − 5400 = 432, and the new cube has side 18 blocks.
In simple words: 5400 lies between 17³ = 4913 and 18³ = 5832: remove 487 blocks or add 432 to make an 18 × 18 × 18 cube.
Teacher's Note:
1. The source answers divided 5400 by 25 (leaving 216 blocks, that is, removing 5184) and added up to 8000 = 20³; these are not the least changes, so they are corrected to 487 removed and side 18.
2. Dividing by a factor is for "smallest divisor" questions; adding or removing blocks means moving to the nearest cube.
3. Check: 17 × 17 × 17 = 4913 and 18 × 18 × 18 = 5832.
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CBSE Class 8 Mathematics Worksheets for Ganita Prakash Part 1 Chapter 01 A Square and A Cube
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