Class 8 Mathematics Practice Assignments: CBSE Class 8 Mathematics Cubes and Cube Roots Practice Paper Assignment Set 07 with Answers (Old NCERT Syllabus)
Review targeted academic assignments with the CBSE Class 8 Mathematics Cubes and Cube Roots Practice Paper Assignment Set 07 with Answers (Old NCERT Syllabus). Built according to official CBSE standards for the 2026-27 term, these downloadable Class 8 Mathematics worksheets support effective daily practice for Chapter 07 Cubes and Cube Roots.
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View or download the dedicated CBSE Class 8 Mathematics Cubes and Cube Roots Practice Paper Assignment Set 07 with Answers (Old NCERT Syllabus) resource below. Engaging with these assignments under focused study conditions ensures continuous academic progress and mastery of the 2026-27 curriculum.
This Class 8 Maths Cubes and Cube Roots practice paper has 20 questions worth 40 marks across Sections A to E: 10 MCQs (two Assertion–Reason), short and long answer questions with internal choices, and two case studies, each solved step by step. It is based on the earlier (old) NCERT Class 8 Maths syllabus and works well for state-board students and for revising these topics. Students practise cubes of numbers, units digits of cubes, cubes of negative numbers, cube roots by prime factorisation and word problems on bundles and cuboids.
Section A: Multiple Choice Questions (1 Mark Each)
(a) 2
(b) 3
(c) 6
(d) 7
Show Answer & Explanation
Answer: (a) 2
Explanation:
1. Start inside: 343 = 7 × 7 × 7, so \( \sqrt[3]{343} \) = 7.
2. Then the expression = \( \sqrt[3]{1 + 7} \) = \( \sqrt[3]{8} \).
3. 8 = 2 × 2 × 2, so \( \sqrt[3]{8} \) = 2.
4. So the value is 2.
Teacher's Note:
1. Work from the innermost root outwards.
2. 7 is the trap: it is only the inner root, not the final answer.
(a) 3
(b) 5
(c) 40
(d) 125
Show Answer & Explanation
Answer: (b) 5
Explanation:
1. 675 = 3 × 3 × 3 × 5 × 5.
2. 3 already forms a triplet; 5 appears only twice.
3. One more 5 completes the triplet: 675 × 5 = 3375 = 15³.
4. So the smallest multiplier is 5.
Teacher's Note:
1. Group the prime factors in threes; multiply by whatever completes each incomplete group.
2. 125 would also give a cube, but it is not the smallest.
(a) 216
(b) 567
(c) 125
(d) 343
Show Answer & Explanation
Answer: (b) 567
Explanation:
1. 216 = 6³, 125 = 5³ and 343 = 7³ are all perfect cubes.
2. 567 = 3 × 3 × 3 × 3 × 7 = 3³ × 3 × 7.
3. The factors 3 (one extra) and 7 are not in triplets, so 567 is not a perfect cube.
4. So the answer is 567.
Teacher's Note:
1. Know the cubes 1³ to 10³: 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000.
2. 567 lies between 8³ = 512 and 9³ = 729, so it cannot be a cube.
(a) 6.327
(b) 63.275
(c) 632.75
(d) 62.757
Show Answer & Explanation
Answer: (d) 62.757
Explanation:
1. 185.193 = \( \frac{185193}{1000} \), so \( \sqrt[3]{185.193} \) = \( \frac{57}{10} \) = 5.7 (since 10³ = 1000).
2. 0.000185193 = \( \frac{185193}{10^9} \), so \( \sqrt[3]{0.000185193} \) = \( \frac{57}{1000} \) = 0.057 (since 1000³ = 10⁹).
3. Sum = 57 + 5.7 + 0.057 = 62.757.
4. So the value is 62.757.
Teacher's Note:
1. The cube root of a decimal has one-third as many decimal places: 3 places → 1 place, 9 places → 3 places.
2. Check: 5.7³ = 185.193.
(a) 22
(b) 12
(c) 11
(d) 13
Show Answer & Explanation
Answer: (c) 11
Explanation:
1. 704 = 2 × 2 × 2 × 2 × 2 × 2 × 11 = 2⁶ × 11.
2. 2⁶ = (2 × 2 × 2) × (2 × 2 × 2) is already a perfect cube; 11 is left over.
3. 704 ÷ 11 = 64 = 4³.
4. So the smallest divisor is 11.
Teacher's Note:
1. Divide out the factors that do not form complete triplets.
2. 22 would leave 32, which is not a cube.
(a) 3375
(b) 3376
(c) 3374
(d) 3275
Show Answer & Explanation
Answer: (a) 3375
Explanation:
1. 15³ = 15 × 15 × 15.
2. 15 × 15 = 225 and 225 × 15 = 3375.
3. So 15³ = 3375.
Teacher's Note:
1. A cube of a number ending in 5 also ends in 5, so 3376 and 3374 are out at once.
2. 225 × 15 = 2250 + 1125 = 3375.
(a) 3
(b) 7
(c) 17
(d) 27
Show Answer & Explanation
Answer: (a) 3
Explanation:
1. The units digit of a cube depends only on the units digit of the number.
2. 3³ = 27 ends in 7.
3. Numbers ending in 7 have cubes ending in 3: 7³ = 343, 17³ = 4913, 27³ = 19683.
4. So the answer is 3.
Teacher's Note:
1. Units-digit pairs for cubes: 3 ↔ 7 and 2 ↔ 8.
2. So a number ending in 7 gives a cube ending in 3, and vice versa.
(a) 2
(b) 3
(c) 4
(d) 9
Show Answer & Explanation
Answer: (b) 3
Explanation:
1. \( \sqrt[3]{64} \) = 4, \( \sqrt[3]{125} \) = 5 and \( \sqrt[3]{27} \) = 3.
2. Value = \( \frac{4 + 5}{3} \) = \( \frac{9}{3} \) = 3.
3. So the value is 3.
Teacher's Note:
1. Find each cube root separately; you cannot add the numbers under the roots.
2. 9 is the trap: it is only the numerator.
Assertion (A): Cube of a negative number is positive.
Reason (R): Odd power of a negative number is always negative.
(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: (d) A is false, but R is true.
Explanation:
1. (−a)³ = (−a) × (−a) × (−a) = −a³, which is negative; e.g. (−2)³ = −8. So A is false.
2. An odd power of a negative number has an odd number of negative factors, so the result is negative. So R is true.
3. So the answer is (d) A is false but R is true.
Teacher's Note:
1. Cube of a negative number is negative; square of a negative number is positive.
2. R actually shows why A is wrong, since 3 is an odd power.
Assertion (A): The one’s digit in the cube root of the cube number 1728 is 6.
Reason (R): The cube root of a number is the factor that we multiply by itself three times to get that number.
(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: (d) A is false, but R is true.
Explanation:
1. 1728 = 12 × 12 × 12, so \( \sqrt[3]{1728} \) = 12, whose one’s digit is 2.
2. Quick rule: a cube ending in 8 has a cube root ending in 2 (2³ = 8). So A is false.
3. R is the correct meaning of a cube root. So R is true.
4. So the answer is (d) A is false but R is true.
Teacher's Note:
1. Units digit pairs for cubes: 2 ↔ 8 and 3 ↔ 7; the digits 0, 1, 4, 5, 6, 9 stay the same.
2. 6³ = 216 ends in 6, not 8, so 6 cannot be the answer.
Section B: Short Answer Questions (2 Marks Each)
Question 11. Find the length of each side of a cube if its volume is 512 cm3.
Answer:
Volume of a cube = side³, so side³ = 512.
512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) = 8 × 8 × 8.
So side = \( \sqrt[3]{512} \) = 8.
Length of each side = 8 cm.
In simple words: The side of a cube is the cube root of its volume, and 8 × 8 × 8 = 512.
Teacher's Note:
1. Group the prime factors in threes and take one factor from each group.
2. Do not divide 512 by 3; that gives nothing useful.
Question 12. What is the smallest number by which 1323 may be multiplied so that the product is a perfect cube?
Answer:
Prime factorisation: 1323 = 3 × 3 × 3 × 7 × 7.
3 forms a triplet (3 × 3 × 3), but 7 occurs only twice.
One more 7 is needed to complete the triplet of 7.
1323 × 7 = 9261 = 21³.
Smallest number to multiply by = 7.
In simple words: 1323 = 3³ × 7², so it needs just one more 7 to become 21³ = 9261.
Teacher's Note:
1. Divide by small primes in order: 1323 ÷ 3 = 441, 441 ÷ 3 = 147, 147 ÷ 3 = 49, 49 = 7 × 7.
2. Check by computing the cube root of the product: 3 × 7 = 21.
Question 13. Evaluate: \( \sqrt[3]{27} + \sqrt[3]{0.008} + \sqrt[3]{0.064} \)
Answer:
\( \sqrt[3]{27} \) = \( \sqrt[3]{3 \times 3 \times 3} \) = 3.
\( \sqrt[3]{0.008} \) = \( \sqrt[3]{0.2 \times 0.2 \times 0.2} \) = 0.2 (as 2³ = 8 and 0.008 has 3 decimal places).
\( \sqrt[3]{0.064} \) = \( \sqrt[3]{0.4 \times 0.4 \times 0.4} \) = 0.4 (as 4³ = 64).
Sum = 3 + 0.2 + 0.4 = 3.6.
In simple words: The three cube roots are 3, 0.2 and 0.4, which add up to 3.6.
Teacher's Note:
1. A decimal with 3 decimal places has a cube root with 1 decimal place.
2. Check: 0.2³ = 0.008 and 0.4³ = 0.064.
Question 14. What is the smallest number by which 1375 should be divided so that the quotient may be a perfect cube?
Answer:
Prime factorisation: 1375 = 5 × 5 × 5 × 11.
5 forms a triplet, but 11 occurs only once and is left over.
Divide by 11 to remove it: 1375 ÷ 11 = 125 = 5³.
Smallest number to divide by = 11.
In simple words: 1375 is 5³ × 11, so dividing by 11 leaves the perfect cube 125.
Teacher's Note:
1. When dividing, remove the factors that are not in complete triplets.
2. When multiplying, add factors to complete the triplets instead.
Section C: Short Answer Questions (3 Marks Each)
Question 15. Sujatha makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many minimum such cuboids will she need to form a cube?
OR
Find the smallest number by which 33275 must be divided so that the quotient has a cube root. Also, find the cube root of the quotient.
Answer:
Volume of one cuboid = 5 × 2 × 5 = 50 cm³ = 2 × 5 × 5.
For a perfect cube every prime factor must occur in triplets.
5 occurs twice (needs one more 5) and 2 occurs once (needs 2 × 2).
So the volume must be multiplied by 5 × 2 × 2 = 20.
20 cuboids give 50 × 20 = 1000 cm³ = 10³, a cube of side 10 cm (2 cuboids along each 5 cm edge and 5 along the 2 cm edge: 2 × 2 × 5 = 20).
Minimum number of cuboids = 20.
OR
Prime factorisation: 33275 = 5 × 5 × 11 × 11 × 11.
Grouping: 5 × 5 × (11 × 11 × 11); 11 forms a triplet but 5 occurs only twice.
Remove the unpaired factors: divide by 5 × 5 = 25.
Quotient = 33275 ÷ 25 = 1331 = 11 × 11 × 11.
Cube root of the quotient = 11.
In simple words: 50 × 20 = 1000 = 10³, so 20 cuboids are needed; in the OR question 33275 ÷ 25 = 1331 = 11³.
Teacher's Note:
1. The number of cuboids is the smallest multiplier that makes the volume a perfect cube.
2. 33275 ends in 5, so start dividing by 5.
Question 16. Find the cube root of \( 4\frac{12}{125} \)
Answer:
Convert to an improper fraction: \( 4\frac{12}{125} = \frac{4 \times 125 + 12}{125} = \frac{512}{125} \).
512 = 8 × 8 × 8 = 8³ and 125 = 5 × 5 × 5 = 5³.
\( \sqrt[3]{\frac{512}{125}} = \frac{\sqrt[3]{512}}{\sqrt[3]{125}} = \frac{8}{5} \).
Cube root = \( \frac{8}{5} \) = \( 1\frac{3}{5} \) = 1.6.
In simple words: 4 12/125 is 512/125, and the cube roots of 512 and 125 are 8 and 5, giving 8/5.
Teacher's Note:
1. Always change a mixed number to an improper fraction before taking a root.
2. Check: (8/5)³ = 512/125 = 4 12/125.
Question 17. Find the smallest number which when multiplied with 3600 will make the product a perfect cube. Find the cube root of the product.
Answer:
Prime factorisation: 3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5.
Grouping in triplets: (2 × 2 × 2) × 2 × (3 × 3) × (5 × 5).
To complete the triplets we need 2 × 2 (for the extra 2), one 3 and one 5.
Smallest multiplier = 2 × 2 × 3 × 5 = 60.
Product = 3600 × 60 = 216000 = (2 × 2 × 2) × (2 × 2 × 2) × (3 × 3 × 3) × (5 × 5 × 5).
Cube root of the product = 2 × 2 × 3 × 5 = 60.
In simple words: 3600 = 2⁴ × 3² × 5² needs 2² × 3 × 5 = 60 more, and 3600 × 60 = 216000 = 60³.
Teacher's Note:
1. Count each prime's power and raise it to the next multiple of 3.
2. Here 2⁴ → 2⁶, 3² → 3³ and 5² → 5³.
Section D: Long Answer Question (5 Marks)
Question 18. Three numbers are in the ratio 1 : 2 : 3 and the sum of their cubes is 4500. Find the numbers.
OR
Find the value of \( \sqrt[3]{968} \times \sqrt[3]{1375} \)
Answer:
Let the numbers be x, 2x and 3x.
x³ + (2x)³ + (3x)³ = 4500, so x³ + 8x³ + 27x³ = 4500.
36x³ = 4500, so x³ = 4500 ÷ 36 = 125.
x = \( \sqrt[3]{125} \) = 5.
The numbers are 5, 10 and 15.
Check: 125 + 1000 + 3375 = 4500.
OR
\( \sqrt[3]{968} \times \sqrt[3]{1375} = \sqrt[3]{968 \times 1375} \).
968 = 2 × 2 × 2 × 11 × 11 and 1375 = 5 × 5 × 5 × 11.
968 × 1375 = (2 × 2 × 2) × (11 × 11 × 11) × (5 × 5 × 5).
Cube root = 2 × 11 × 5 = 110.
In simple words: 36x³ = 4500 gives x = 5, so the numbers are 5, 10, 15; in the OR question the value is 110.
Teacher's Note:
1. (2x)³ = 8x³, not 2x³: cube the number and the variable.
2. The product of cube roots is the cube root of the product.
Section E: Case Study Based Questions (4 Marks Each)
Question 19. Pencil Packing: A pencil factory packs pencils in bundles, each containing the same number of pencils, forming a cubic number total. For a bulk supply to a school, the manager counts and finds that each bundle contains exactly 1331 pencils. He is curious if this number is a perfect cube, wants to know the cube root, and also wants to calculate the total pencils if 27 such bundles are supplied.
(a) Verify whether 1331 is a perfect cube.
(b) Find the cube root of 1331.
(c) Find the total pencils in 27 bundles.
Answer:
(a) 1331 ÷ 11 = 121, 121 ÷ 11 = 11, so 1331 = 11 × 11 × 11.
The factor 11 forms one complete triplet, so 1331 is a perfect cube (1331 = 11³).
(b) \( \sqrt[3]{1331} \) = 11.
(c) Total pencils = 27 × 1331 = 35937.
Note: 27 × 1331 = 3³ × 11³ = 33³, so the total is again a perfect cube.
In simple words: 1331 = 11³, so its cube root is 11, and 27 bundles hold 27 × 1331 = 35937 pencils.
Teacher's Note:
1. 1331 is divisible by 11 (alternate digit sums 1 + 3 and 3 + 1 are equal).
2. The product of two perfect cubes is always a perfect cube.
Question 20. Mohan has to prepare a physics project in form of a cubical box for a social work campaign but he had a cuboidal box of sides 4 cm, 2 cm, 4 cm. Now he has to change it in the form of cube so that he can complete his project. For this, he needed more cuboids so that he can make his project in form of cube.
(a) What is the volume of the cuboidal box? [1]
(b) Find value of (0.02)3. [1]
(c) How many cuboids are more needed? [2]
Answer:
(a) Volume of the cuboid = l × b × h = 4 × 2 × 4 = 32 cm³.
(b) \( (0.02)^3 = \frac{2}{100} \times \frac{2}{100} \times \frac{2}{100} = \frac{8}{1000000} \) = 0.000008.
(c) 32 = 2 × 2 × 2 × 2 × 2 = 2⁵. For a perfect cube the power of 2 must be a multiple of 3, so we need 2⁶ = 64.
So the volume must be doubled: 32 × 2 = 64 cm³ = 4³, a cube of side 4 cm.
Two cuboids of 4 cm × 4 cm × 2 cm placed one on top of the other (2 cm + 2 cm = 4 cm) make a 4 cm × 4 cm × 4 cm cube.
Mohan already has 1 cuboid, so he needs 1 more cuboid (2 cuboids in all).
In simple words: Two 4 × 4 × 2 boxes stacked make a 4 cm cube, so Mohan needs just one more box.
Teacher's Note:
1. Find the smallest multiplier that makes the volume a perfect cube; that is the number of cuboids.
2. Always check that the cuboids really fit together: two 2 cm heights make 4 cm.
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