NCERT Solutions for class 8 Maths Chapter 7 - Cubes and Cube Roots Exercise 7.1

NCERT Solutions for Class 8 Maths Chapter 7- Cubes and Cube Roots Exercise 7.1 have been provided here for students to prepare well for the exam. These solutions have been prepared by the subject experts at BYJU’S and are in accordance with the NCERT syllabus and guidelines. The steps given in the examples are followed while providing the NCERT solutions for the questions present in the exercises.

These solutions are helpful for students to score well in the examination, as it works for them as a reference tool while working on the revision. The NCERT Class 8 Maths Solutions materials are in the context of the upcoming session of 2023-2024.

NCERT Solutions for Class 8 Maths Chapter 7 – Cubes and Cube Roots Exercise 7.1

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Exercise 7.2 Solutions 3 Questions (2 Long Answer Questions, 1 Short Answer Question)

Access Answers of Maths NCERT Class 8 Chapter 7 – Cubes and Cube Roots Exercise 7.1 Page Number 114

Exercise 7.1 Page: 114

1. Which of the following numbers are not perfect cubes?

(i) 216

Solution:

By resolving 216 into prime factor,

216 = 2×2×2×3×3×3

By grouping the factors in triplets of equal factors, 216 = (2×2×2)×(3×3×3)

Here, 216 can be grouped into triplets of equal factors,

∴ 216 = (2×3) = 6

Hence, 216 is cube of 6.

(ii) 128

Solution:

By resolving 128 into prime factor,

128 = 2×2×2×2×2×2×2

By grouping the factors in triplets of equal factors, 128 = (2×2×2)×(2×2×2)×2

Here, 128 cannot be grouped into triplets of equal factors, we are left of with one factors 2 .

∴ 128 is not a perfect cube.

(iii) 1000 Solution:

By resolving 1000 into prime factor,

1000 = 2×2×2×5×5×5

By grouping the factors in triplets of equal factors, 1000 = (2×2×2)×(5×5×5)

Here, 1000 can be grouped into triplets of equal factors,

∴ 1000 = (2×5) = 10

Hence, 1000 is cube of 10.

(iv) 100 Solution:

By resolving 100 into prime factor,

100 = 2×2×5×5

Here, 100 cannot be grouped into triplets of equal factors.

∴ 100 is not a perfect cube.

(v) 46656

Solution:

By resolving 46656 into prime factor,

46656 = 2×2×2×2×2×2×3×3×3×3×3×3

By grouping the factors in triplets of equal factors, 46656 = (2×2×2)×(2×2×2)×(3×3×3)×(3×3×3)

Here, 46656 can be grouped into triplets of equal factors,

∴ 46656 = (2×2×3×3) = 36

Hence, 46656 is cube of 36.

2. Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube.

(i) 243

Solution:

By resolving 243 into prime factor,

243 = 3×3×3×3×3

By grouping the factors in triplets of equal factors, 243 = (3×3×3)×3×3

Here, 3 cannot be grouped into triplets of equal factors.

∴ We will multiply 243 by 3 to get perfect cube.

(ii) 256 Solution:

By resolving 256 into prime factor,

256 = 2×2×2×2×2×2×2×2

By grouping the factors in triplets of equal factors, 256 = (2×2×2)×(2×2×2)×2×2

Here, 2 cannot be grouped into triplets of equal factors.

∴ We will multiply 256 by 2 to get perfect cube.

(iii) 72

Solution:

By resolving 72 into prime factor,

72 = 2×2×2×3×3

By grouping the factors in triplets of equal factors, 72 = (2×2×2)×3×3

Here, 3 cannot be grouped into triplets of equal factors.

∴ We will multiply 72 by 3 to get perfect cube.

(iv) 675 Solution:

By resolving 675 into prime factor,

675 = 3×3×3×5×5

By grouping the factors in triplets of equal factors, 675 = (3×3×3)×5×5

Here, 5 cannot be grouped into triplets of equal factors.

∴ We will multiply 675 by 5 to get perfect cube.

(v) 100 Solution:

By resolving 100 into prime factor,

100 = 2×2×5×5

Here, 2 and 5 cannot be grouped into triplets of equal factors.

∴ We will multiply 100 by (2×5) 10 to get perfect cube.

3. Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube.

(i) 81

Solution:

By resolving 81 into prime factor,

81 = 3×3×3×3

By grouping the factors in triplets of equal factors, 81 = (3×3×3)×3

Here, 3 cannot be grouped into triplets of equal factors.

∴ We will divide 81 by 3 to get perfect cube.

(ii) 128 Solution:

By resolving 128 into prime factor,

128 = 2×2×2×2×2×2×2

By grouping the factors in triplets of equal factors, 128 = (2×2×2)×(2×2×2)×2

Here, 2 cannot be grouped into triplets of equal factors.

∴ We will divide 128 by 2 to get perfect cube.

(iii) 135 Solution:

By resolving 135 into prime factor,

135 = 3×3×3×5

By grouping the factors in triplets of equal factors, 135 = (3×3×3)×5

Here, 5 cannot be grouped into triplets of equal factors.

∴ We will divide 135 by 5 to get perfect cube.

(iv) 192 Solution:

By resolving 192 into prime factor,

192 = 2×2×2×2×2×2×3

By grouping the factors in triplets of equal factors, 192 = (2×2×2)×(2×2×2)×3

Here, 3 cannot be grouped into triplets of equal factors.

∴ We will divide 192 by 3 to get perfect cube.

(v) 704 Solution:

By resolving 704 into prime factor,

704 = 2×2×2×2×2×2×11

By grouping the factors in triplets of equal factors, 704 = (2×2×2)×(2×2×2)×11

Here, 11 cannot be grouped into triplets of equal factors.

∴ We will divide 704 by 11 to get perfect cube.

4. Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube?

Solution:

Given, side of cube is 5 cm, 2 cm and 5 cm.

∴ Volume of cube = 5×2×5 = 50

50 = 2×5×5

Here, 2 , 5 and 5 cannot be grouped into triplets of equal factors.

∴ We will multiply 50 by (2×2×5) 20 to get perfect cube. Hence, 20 cuboid is needed.

The concepts of cubes and perfect cubes are the base of Exercise 7.1, Class 8 NCERT Maths Chapter 7, Cubes and Cube Roots . There are 4 main questions in this exercise. The first three questions contain five sub-questions each. The exercise asks questions related to finding out if a number is a perfect cube or not. In order to find this, a strong knowledge of the concept of prime factorisation is mandatory. Solving this exercise will help students to get a strong grip on the concept of cubes and perfect cubes.

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NCERT Solutions for Class 8