NCERT Solutions for Class 8 Maths Chapter 7: Cubes and Cube Roots
The provided NCERT Solutions for Class 8 Maths Chapter 7 on Cubes and Cube Roots offer a comprehensive analysis of the questions within the chapter, aligning with the principles outlined in the Class 8 NCERT Textbook.
Access Answers to NCERT Solutions for Class 8 Maths Chapter 7: Cubes and Cube Roots
Students can access the NCERT Solutions for Class 8 Maths Chapter 7: Cubes and Cube Roots. Curated by experts according to the CBSE syllabus for 2023–2024, these step-by-step solutions make Maths much easier to understand and learn for the students. These solutions can be used in practice by students to attain skills in solving problems, reinforce important learning objectives, and be well-prepared for tests.
Exercise 7.1
Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube.
(i) 243
(ii) 256
(iii) 72
(iv) 675
(v) 100
(i) By resolving 243 into a 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 the perfect cube.
(ii) By resolving 256 into a 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 the perfect cube.
(iii) By resolving 72 into a 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 the perfect cube.
(iv) By resolving 675 into a 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 the perfect cube.
(v) By resolving 100 into a 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 the perfect cube.
Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube.
(i) 81
(ii) 128
(iii) 135
(iv) 192
(v) 704
(i) By resolving 81 into a 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 the perfect cube.
(ii) By resolving 128 into a 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 the perfect cube.
(iii) 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 the perfect cube.
(iv) By resolving 192 into a 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 the perfect cube.
(v) By resolving 704 into a 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 the perfect cube.
Parikshit makes a cuboid of plasticine with sides 5 cm, 2 cm, and 5 cm. How many such cuboids will he need to form a cube?
Given the sides of the cube are 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 the perfect cube. Hence, 20 cuboids are 1.
Which of the following numbers are not perfect cubes?
(i) 216
(ii) 128
(iii) 1000
(iv) 100
(v) 46656
(i) By resolving 216 into a 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 the cube of 6.
(ii) By resolving 128 into a 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, and we are left with one factor: 2.
∴ 128 is not a perfect cube.
(iii) 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 the cube of 10.
(iv) By resolving 100 into a prime factor,
100 = 2×2×5×5
Here, 100 cannot be grouped into triplets of equal factors.
∴ 100 is not a perfect cube.
(v) 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 the cube of 36.
Exercise 7.2
You are told that 1,331 is a perfect cube. Can you guess without factorisation what its cube root is? Similarly, guess the cube roots of 4913, 12167, and 32768.
(i) By grouping the digits, we get 1 and 331
We know that since the unit digit of the cube is 1, the unit digit of the cube root is 1.
∴ We get 1 as the unit digit of the cube root of 1331.
The cube of 1 matches the number of the second group.
∴ The ten’s digit of our cube root is taken as the unit place of the smallest number.
We know that the unit’s digit of the cube of a number having digit as unit’s place 1 is 1.
∴ ∛1331 = 11
(ii) By grouping the digits, we get 4 and 913
We know that since the unit digit of the cube is 3, the unit digit of the cube root is 7.
∴ we get 7 as the unit digit of the cube root of 4913. We know 13 = 1 and 23 = 8 , 1 > 4 > 8
Thus, 1 is taken as the tens digit of the cube root.
∴ ∛4913 = 17
(iii) By grouping the digits, we get 12 and 167.
We know that since the unit digit of the cube is 7, the unit digit of the cube root is 3.
∴ 3 is the unit digit of the cube root of 12167 We know 23 = 8 and 33 = 27 , 8 > 12 > 27
Thus, 2 is taken as the tens digit of the cube root.
∴ ∛12167= 23
(iv) By grouping the digits, we get 32 and 768.
We know that since the unit digit of the cube is 8, the unit digit of the cube root is 2.
∴ 2 is the unit digit of the cube root of 32768. We know 33 = 27 and 43 = 64 , 27 > 32 > 64
Thus, 3 is taken as the tens digit of the cube root.
∴ ∛32768= 32
Find the cube root of each of the following numbers by the prime factorisation method.
(i) 64
(ii) 512
(iii) 10648
(iv) 27000
(v) 15625
(vi) 13824
(vii) 110592
(viii) 46656
(ix) 175616
(x) 91125
(i) 64 = 2×2×2×2×2×2
By grouping the factors in triplets of equal factors, 64 = (2×2×2)×(2×2×2)
Here, 64 can be grouped into triplets of equal factors.
∴ 64 = 2×2 = 4
Hence, 4 is the cube root of 64.
(ii) 512 = 2×2×2×2×2×2×2×2×2
By grouping the factors in triplets of equal factors, 512 = (2×2×2)×(2×2×2)×(2×2×2)
Here, 512 can be grouped into triplets of equal factors.
∴ 512 = 2×2×2 = 8
Hence, 8 is the cube root of 512.
(iii) 10648 = 2×2×2×11×11×11
By grouping the factors in triplets of equal factors, 10648 = (2×2×2)×(11×11×11)
Here, 10648 can be grouped into triplets of equal factors.
∴ 10648 = 2 ×11 = 22
Hence, 22 is the cube root of 10648.
(iv) 27000 = 2×2×2×3×3×3×3×5×5×5
By grouping the factors in triplets of equal factors, 27000 = (2×2×2)×(3×3×3)×(5×5×5)
Here, 27000 can be grouped into triplets of equal factors.
∴ 27000 = (2×3×5) = 30
Hence, 30 is the cube root of 27000.
(v) 15625 = 5×5×5×5×5×5
By grouping the factors in triplets of equal factors, 15625 = (5×5×5)×(5×5×5)
Here, 15625 can be grouped into triplets of equal factors.
∴ 15625 = (5×5) = 25
Hence, 25 is the cube root of 15625.
(vi) 13824 = 2×2×2×2×2×2×2×2×2×3×3×3
By grouping the factors in triplets of equal factors,
13824 = (2×2×2)×(2×2×2)×(2×2×2)×(3×3×3)
Here, 13824 can be grouped into triplets of equal factors.
∴ 13824 = (2×2× 2×3) = 24
Hence, 24 is the cube root of 13824.
(vii) 110592 = 2×2×2×2×2×2×2×2×2×2×2×2×3×3×3
By grouping the factors in triplets of equal factors,
110592 = (2×2×2)×(2×2×2)×(2×2×2)×(2×2×2)×(3×3×3)
Here, 110592 can be grouped into triplets of equal factors.
∴ 110592 = (2×2×2×2 × 3) = 48
Hence, 48 is the cube root of 110592.
(viii) 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, 36 is the cube root of 46656.
(ix) 175616 = 2×2×2×2×2×2×2×2×2×7×7×7
By grouping the factors in triplets of equal factors,
175616 = (2×2×2)×(2×2×2)×(2×2×2)×(7×7×7)
Here, 175616 can be grouped into triplets of equal factors.
∴ 175616 = (2×2×2×7) = 56
Hence, 56 is the cube root of 175616.
(x) 91125 = 3×3×3×3×3×3×3×5×5×5
By grouping the factors in triplets of equal factors, 91125 = (3×3×3)×(3×3×3)×(5×5×5)
Here, 91125 can be grouped into triplets of equal factors.
∴ 91125 = (3×3×5) = 45
Hence, 45 is the cube root of 91125.
State true or false.
(i) Cube of any odd number is even.
(ii) A perfect cube does not end with two zeros.
(iii) If the cube of a number ends with 5, then its cube ends with 25.
(iv) There is no perfect cube which ends with 8.
(v) The cube of a two-digit number may be a three-digit number.
(vi) The cube of a two-digit number may have seven or more digits.
(vii) The cube of a single-digit number may be a single-digit number.
(i) False
(ii) True
(iii) False
(iv) False
(v) False
(vi) False
(vii) True
Frequently Asked Questions
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