Class 8 Mathematics Practice Assignments: CBSE Class 8 Mathematics A Square and A Cube Practice Paper Assignment Set 01 with Answers
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Download Ganita Prakash Part 1 Chapter 01 A Square and A Cube Assignment PDF with Solutions
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This is Practice Paper 06 for CBSE Class 8 Maths Ganita Prakash Chapter 1 A Square and A Cube: 40 marks, 20 questions across Sections A to E, each followed by a worked answer. The questions focus on reasoning with factors, such as why only square-numbered lockers stay open, how many factors the square of a prime has, end zeros of squares and cubes, cubes as sums of odd numbers, LCM-based square grids, and a real-life problem of cutting 7 cm dice from a 10648 cm³ wooden cube.
Section A: Multiple Choice Questions (1 Mark Each)
(a) It is a prime number.
(b) It has exactly three factors.
(c) It is a perfect square.
(d) It is a perfect cube.
Show Answer & Explanation
Answer: (c) It is a perfect square.
Explanation:
1. Locker n is toggled once by every person whose number is a factor of n, so the number of toggles = number of factors of n.
2. Factors pair up (a × b = n), except when a number is multiplied by itself, which happens only for perfect squares.
3. So an odd number of toggles means an odd number of factors, which means the locker number is a perfect square.
Teacher's Note:
1. A prime has 2 factors (even), so (a) is wrong.
2. "Exactly three factors" fits only squares of primes such as 4, 9, 25, but 16 is toggled 5 times, so (b) need not be true.
(a) Yes, always, because the number of zeros is even.
(b) Yes, if the non-zero part of the number is also a perfect square.
(c) No, because perfect squares cannot end with 8 zeros.
(d) Cannot be determined without more information.
Show Answer & Explanation
Answer: (b) Yes, if the non-zero part of the number is also a perfect square.
Explanation:
1. A number ending with 8 zeros can be written as N × 10⁸, where N is the non-zero part.
2. 10⁸ = (10⁴)² is a perfect square, so N × 10⁸ is a perfect square exactly when N is a perfect square.
3. For example 4 × 10⁸ = (2 × 10⁴)² is a square, but 2 × 10⁸ is not.
4. So the correct statement is (b) Yes, if the non-zero part of the number is also a perfect square.
Teacher's Note:
1. An even number of end zeros is needed for a square, but it is not enough on its own.
2. Option (a) originally read "Yes, because the number of zeros is even"; it has been made "Yes, always" so that only (b) is fully correct.
(a) 2
(b) 3
(c) 4
(d) An even number
Show Answer & Explanation
Answer: (b) 3
Explanation:
1. If x is a prime p, its only factors are 1 and p.
2. The factors of p² are 1, p and p².
3. So x² has 3 factors. For example, 25 = 5² has factors 1, 5, 25.
Teacher's Note:
1. The square of a prime always has exactly three factors.
2. Being a square, it must have an odd number of factors, so (d) is ruled out at once.
(a) N cannot be a perfect square.
(b) N must have an odd number of factors.
(c) N is a perfect square because 12 is an even number.
(d) The number of factors of N is irrelevant to whether it is a perfect square.
Show Answer & Explanation
Answer: (a) N cannot be a perfect square.
Explanation:
1. A perfect square always has an odd number of factors, because its square root pairs with itself.
2. N has 12 factors, which is an even number.
3. So N cannot be a perfect square.
Teacher's Note:
1. The original stem said "If N is a perfect square", which contradicts 12 factors; that phrase is removed, and option (c) is reworded so that only one option is true.
2. Example: 60 has 12 factors and is not a square.
(a) 4
(b) 6
(c) 8
(d) 2
Show Answer & Explanation
Answer: (b) 6
Explanation:
1. The units digit of a square depends only on the units digit of the number.
2. 34 ends in 4, and 4² = 16 ends in 6.
3. So 34² ends in 6 (indeed 34² = 1156).
Teacher's Note:
1. Numbers ending in 4 or 6 have squares ending in 6.
2. Look only at the last digit to answer such questions quickly.
(a) They must appear in pairs.
(b) They must appear in groups of three.
(c) They must appear an even number of times.
(d) They must all be different.
Show Answer & Explanation
Answer: (b) They must appear in groups of three.
Explanation:
1. The number is n × n × n = n³, a perfect cube.
2. Each prime factor of n appears once in each of the three copies of n.
3. So in n³ every prime factor appears in groups of three; e.g. 6³ = (2 × 2 × 2) × (3 × 3 × 3).
Teacher's Note:
1. The original options (b) "must appear three times" and (d) "must be prime numbers" are reworded: prime factors are always primes, and a factor may appear 6 or 9 times (as in 4³ = 2⁶).
2. Pairs are for squares, triplets are for cubes.
(a) A number ending with 8.
(b) A number ending with 7.
(c) A number ending with exactly 2 zeroes.
(d) A number ending with 9.
Show Answer & Explanation
Answer: (c) A number ending with exactly 2 zeroes.
Explanation:
1. Cubes can end in any digit: 2³ = 8, 3³ = 27, 9³ = 729, so (a), (b) and (d) are possible.
2. Zeros at the end of a cube come in groups of three (10³ = 1000, 100³ = 1000000).
3. So a number ending with exactly 2 zeroes cannot be a cube.
Teacher's Note:
1. Unlike squares, cubes can end in every digit 0 to 9.
2. The word "exactly" is added, because 1000 also ends in two zeros.
(a) The sum is 4³.
(b) The sum is a perfect square.
(c) The sum is 5³.
(d) The sum is 216.
Show Answer & Explanation
Answer: (c) The sum is 5³.
Explanation:
1. The odd numbers can be grouped as 1 | 3 + 5 | 7 + 9 + 11 | 13 + 15 + 17 + 19 | 21 + 23 + 25 + 27 + 29 | …, and each group adds up to a cube.
2. The 5th group has 5 numbers: 21 + 23 + 25 + 27 + 29 = 125.
3. 125 = 5 × 5 × 5, so the sum is 5³.
Teacher's Note:
1. The n-th group has n odd numbers and adds up to n³.
2. 125 is not a perfect square (11² = 121, 12² = 144), so (b) is wrong.
Assertion (A): The number 1156 is a perfect square.
Reason (R): The prime factorization of 1156 can be grouped into two identical sets of factors.
(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. 1156 = 2 × 2 × 17 × 17 = (2 × 17) × (2 × 17) = 34 × 34.
2. So 1156 = 34² is a perfect square. A is true.
3. The prime factors split into two identical sets {2, 17} and {2, 17}, so R is true.
4. Splitting into two identical sets is exactly what makes a number a square, so R explains A. Answer: (a).
Teacher's Note:
1. Two identical sets means every prime factor occurs in pairs.
2. Check: 34 × 34 = 1156.
Assertion (A): The number 500 is not a perfect cube.
Reason (R): The prime factorization of 500 is 2 × 2 × 5 × 5 × 5, and the factors cannot be split into three identical groups.
(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. 500 = 2 × 2 × 5 × 5 × 5; the three 5s form a triplet but there are only two 2s.
2. So the factors cannot be split into three identical groups, and R is true.
3. Hence 500 is not a perfect cube (7³ = 343 < 500 < 512 = 8³). A is true.
4. R gives the reason for A. Answer: (a).
Teacher's Note:
1. Multiplying 500 by 2 gives 1000 = 10³.
2. A perfect cube's prime factors can always be split into three identical groups.
Section B: Short Answer Questions (2 Marks Each)
Question 11. Find the square root of 1296 by the prime factorisation method.
Answer:
1296 ÷ 2 = 648, ÷ 2 = 324, ÷ 2 = 162, ÷ 2 = 81, ÷ 3 = 27, ÷ 3 = 9, ÷ 3 = 3, ÷ 3 = 1.
1296 = (2 × 2) × (2 × 2) × (3 × 3) × (3 × 3).
Take one factor from each pair: √1296 = 2 × 2 × 3 × 3 = 36.
In simple words: Pair up the prime factors of 1296 and take one from each pair: 36.
Teacher's Note:
1. Check: 36 × 36 = 1296.
2. 1296 ends in 6, so its square root ends in 4 or 6; this agrees with 36.
Question 12. Find the cube root of 2197 by the prime factorisation method.
Answer:
2197 is not divisible by 2, 3, 5, 7 or 11. Try 13: 2197 ÷ 13 = 169, 169 ÷ 13 = 13, 13 ÷ 13 = 1.
2197 = 13 × 13 × 13, one complete triplet.
So ∛2197 = 13.
In simple words: 2197 is three 13s multiplied together, so its cube root is 13.
Teacher's Note:
1. When small primes do not divide, keep trying the next primes.
2. Check: 13 × 13 = 169 and 169 × 13 = 2197.
Question 13. Find the smallest square number which is exactly divisible by 6, 8, and 15.
Answer:
6 = 2 × 3, 8 = 2 × 2 × 2, 15 = 3 × 5, so LCM = 2 × 2 × 2 × 3 × 5 = 120.
In 120 = (2 × 2) × 2 × 3 × 5, the factors 2, 3 and 5 are unpaired.
Multiply by 2 × 3 × 5 = 30: 120 × 30 = 3600 = 60².
The smallest square number is 3600.
In simple words: Find the LCM 120 and multiply by 30 to pair up every factor, giving 3600.
Teacher's Note:
1. Check: 3600 ÷ 6 = 600, 3600 ÷ 8 = 450, 3600 ÷ 15 = 240.
2. Only one extra 2 is needed because 120 already has a pair of 2s.
Question 14. What is the smallest number by which 384 must be multiplied to get a perfect cube?
Answer:
384 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 (seven 2s and one 3).
Six 2s make two triplets; one 2 is left, so two more 2s are needed. The single 3 needs two more 3s.
Multiply by 2 × 2 × 3 × 3 = 36: 384 × 36 = 13824 = 24³.
The smallest multiplier is 36.
In simple words: 384 has a spare 2 and a spare 3; complete both triplets by multiplying by 2 × 2 × 3 × 3 = 36.
Teacher's Note:
1. Check: 24³ = 576 × 24 = 13824.
2. Each leftover factor needs enough copies to reach a group of three.
Section C: Short Answer Questions (3 Marks Each)
Question 15. Find the smallest number by which 1575 must be divided so that it becomes a perfect square. Also, find the square root of the quotient.
Answer:
1575 = 3 × 3 × 5 × 5 × 7 = (3 × 3) × (5 × 5) × 7.
The factor 7 is unpaired, so divide by 7: 1575 ÷ 7 = 225.
225 = (3 × 3) × (5 × 5), so √225 = 3 × 5 = 15.
The smallest divisor is 7 and the square root of the quotient is 15.
In simple words: Divide out the lonely 7 to get 225, whose square root is 15.
Teacher's Note:
1. 1575 ends in 5, so start dividing by 5 or 3; both work.
2. Check: 15 × 15 = 225 and 225 × 7 = 1575.
Question 16. Find the smallest number which when multiplied with 1600 will make the product a perfect cube. Find the cube root of the product.
Answer:
1600 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 (six 2s and two 5s).
The six 2s make two triplets; the 5s need one more 5.
Multiply by 5: 1600 × 5 = 8000.
8000 = (2 × 2 × 2) × (2 × 2 × 2) × (5 × 5 × 5), so ∛8000 = 2 × 2 × 5 = 20.
The smallest multiplier is 5 and the cube root of the product is 20.
In simple words: Add one more 5 to make 8000, which is 20³.
Teacher's Note:
1. 8000 = 8 × 1000 = 2³ × 10³ = 20³, a quick check.
2. 1600 is a perfect square (40²) but not a cube; the two ideas are different.
Question 17. Find the smallest number by which 2048 must be divided so that the quotient has a cube root. Also, find the cube root of the quotient.
Answer:
2048 = 2 multiplied by itself 11 times.
Eleven 2s make three triplets (nine 2s) with two 2s left over.
Divide by 2 × 2 = 4: 2048 ÷ 4 = 512.
512 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2), so ∛512 = 2 × 2 × 2 = 8.
The smallest divisor is 4 and the cube root of the quotient is 8.
In simple words: Remove the two extra 2s by dividing by 4; 512 = 8³ remains.
Teacher's Note:
1. Dividing by just 2 leaves ten 2s, which still is not a multiple of 3.
2. Check: 8³ = 512 and 512 × 4 = 2048.
Section D: Long Answer Question (5 Marks)
Question 18. A group of students can be arranged into rows of 18, 24, or 30 without any student being left out. What is the minimum number of students if they can also be arranged in a perfect square grid?
Answer:
The number of students must be a multiple of 18, 24 and 30, so it is a multiple of their LCM.
18 = 2 × 3 × 3, 24 = 2 × 2 × 2 × 3, 30 = 2 × 3 × 5.
LCM = 2 × 2 × 2 × 3 × 3 × 5 = 360.
For a perfect square grid the number must also be a perfect square.
360 = (2 × 2) × 2 × (3 × 3) × 5; the factors 2 and 5 are unpaired.
Multiply by 2 × 5 = 10: 360 × 10 = 3600 = 60².
The minimum number of students is 3600 (a grid of 60 rows × 60 columns).
In simple words: Take the LCM 360 of 18, 24 and 30 and multiply by 10 to make it a square: 3600.
Teacher's Note:
1. The answer must be a multiple of the LCM and a square at the same time.
2. Check: 3600 ÷ 18 = 200, 3600 ÷ 24 = 150, 3600 ÷ 30 = 120.
3. Multiplying by only the unpaired primes keeps the answer as small as possible.
Section E: Case Study Based Questions (4 Marks Each)
Question 19. Case Study 1 (Locker Puzzle Extended): In a palace, 400 lockers are arranged. The same process is followed as in Queen Ratnamanjuri’s puzzle: Person 1 toggles every locker, Person 2 toggles every 2nd, Person 3 toggles every 3rd, and so on, until all 400 have had their turn.
Raghav observed that only lockers numbered with perfect squares would remain open. He wanted to find patterns in their distribution.
Answer the following:
(a) How many lockers remain open?
(b) Write the numbers of the first five lockers that remain open.
(c) Find the largest locker number that remains open.
(d) Explain why locker number 289 is open.
Answer:
Open lockers are those with perfect-square numbers.
(a) The squares from 1 to 400 are 1², 2², …, 20² (since 20² = 400). So 20 lockers remain open.
(b) The first five open lockers are 1, 4, 9, 16 and 25.
(c) The largest square not more than 400 is 20² = 400, so locker 400 is the largest open one.
(d) 289 = 17 × 17. Its factors are 1, 17 and 289, which is 3 factors (an odd number).
So locker 289 is toggled 3 times (by persons 1, 17 and 289): open, closed, open. It ends up open.
In simple words: Only square-numbered lockers stay open; there are 20 of them up to 400.
Teacher's Note:
1. Number of toggles of a locker = number of factors of its number.
2. For (d) it is not enough to say "289 is a square"; show why a square has an odd number of factors.
Question 20. Case Study 2 (Cube Roots in Real Life): An artisan was tasked with crafting wooden dice for a new board game. Starting with a large wooden cube of 10,648 cm³, he planned to cut it into smaller cubical dice, each with an edge of 7 cm. He was curious about how many dice he could make and if there would be any wood left over from the original cube. This calculation was a critical first step in his design, as it would determine the final number of dice for the game.
Answer the following:
(a) Verify whether 10648 is a perfect cube.
(b) Find the cube root of 10648.
(c) How many dice of side 7 cm can be cut out from the big cube?
(d) If each die is painted at ₹2 per cm² of surface area, find the cost of painting one die.
Answer:
(a) 10648 = 2 × 2 × 2 × 11 × 11 × 11 = (2 × 2 × 2) × (11 × 11 × 11).
Every prime factor occurs in a triplet, so 10648 is a perfect cube.
(b) Taking one factor from each triplet, ∛10648 = 2 × 11 = 22. So the big cube has an edge of 22 cm.
(c) Along each edge of 22 cm we can fit 3 dice of 7 cm (3 × 7 = 21 cm), with 1 cm left over. Top view of one layer:
So the dice fit as 3 rows × 3 columns × 3 layers = 27 dice.
(Dividing volumes, 10648 ÷ 343 ≈ 31, only gives an upper limit; a cube cannot actually be cut into 31 whole 7 cm dice, because the 1 cm strips left over cannot be joined into dice.)
Wood used = 27 × 343 = 9261 cm³, so 10648 − 9261 = 1387 cm³ of wood is left over.
(d) A die has 6 square faces, each of area 7 × 7 = 49 cm².
Surface area = 6 × 49 = 294 cm². Cost = 294 × ₹2 = ₹588.
In simple words: The big cube has edge 22 cm, so only 3 dice fit along each edge, giving 3 × 3 × 3 = 27 dice.
Teacher's Note:
1. The source answer gave 31 dice by dividing volumes; that ignores the shape, and the correct count of whole dice is 27.
2. Always check how many pieces fit along each edge when cutting a cube into smaller cubes.
3. Surface area of a cube = 6 × side².
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CBSE Class 8 Mathematics Assignments for Ganita Prakash Part 1 Chapter 01 A Square and A Cube
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