NCERT Solutions for Class 8 Maths Maths - NCERT Solution Chapter 7

Cubes and Cube Roots Class 8

Chapter 7 Cubes and Cube Roots Exercise 7.1, 7.2 Solutions

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Exercise 7.1 : Solutions of Questions on Page Number : 114

Q1 :  

Which of the following numbers are notperfect cubes?

(i) 216

(ii) 128

(iii) 1000

(iv) 100

(v) 46656


Answer :

(i) The prime factorisation of 216 is as follows.

2

216

2

108

2

54

3

27

3

9

3

3

1

216 = 2 × 2 × 2 × 3 × 3 × 3 = 23 × 33

Here, as each prime factor is appearing as many times as a perfect multiple of 3, therefore, 216 is a perfect cube.

(ii)The prime factorisation of 128 is as follows.

2

128

2

64

2

32

2

16

2

8

2

4

2

2

1

128 = 2 × 2 × 2 × 2 × 2 × 2 × 2

Here, each prime factor is not appearing as many times as a perfect multiple of 3. One 2 is remaining after grouping the triplets of 2. Therefore, 128 is not a perfect cube.

(iii) The prime factorisation of 1000 is as follows.

2

1000

2

500

2

250

5

125

5

25

5

5

1

1000 = 2 × 2 × 2 × 5 × 5 × 5

Here, as each prime factor is appearing as many times as a perfect multiple of 3, therefore, 1000 is a perfect cube.

(iv)The prime factorisation of 100 is as follows.

2

100

2

50

5

25

5

5

1

100 = 2 × 2 × 5 × 5

Here, each prime factor is not appearing as many times as a perfect multiple of 3. Two 2s and two 5s are remaining after grouping the triplets. Therefore, 100 is not a perfect cube.

(v)The prime factorisation of 46656 is as follows.

2

46656

2

23328

2

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Q2 :  

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


Answer :

(i) 243 = 3 x 3 x 3 x 3 x 3

Here, two 3s are left which are not in a triplet. To make 243 a cube, one more 3 is required.

In that case, 243 x 3 = 3 x 3 x 3 x 3 x 3 x 3 = 729 is a perfect cube.

Hence, the smallest natural number by which 243 should be multiplied to make it a perfect cube is 3.

(ii) 256 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2

Here, two 2s are left which are not in a triplet. To make 256 a cube, one more 2 is required.

Then, we obtain

256 x 2 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 512 is a perfect cube.

Hence, the smallest natural number by which 256 should be multiplied to make it a perfect cube is 2.

(iii) 72 = 2 x 2 x 2 x 3 x 3

Here, two 3s are left which are not in a triplet. To make 72 a cube, one more 3 is required.

Then, we obtain

72 x 3 = 2 x 2 x 2 x 3 x 3 x 3 = 216 is a perfect cube.

Hence, the smallest natural number by which 72 should be multiplied to make it a perfect cube is 3.

(iv) 675 = 3 x 3 x 3 x 5 x 5

Here, two 5s are left which are not in a triplet. To make 675 a cube, one more 5 is required.

Then, we obtain

675 x 5 = 3 x 3 x 3 x 5 x 5 x 5 = 3375 is a perfect cube.

Hence, the smallest natural number by which 675 should be multiplied to make it a perfect cube is 5.

(v) 100 = 2 x 2 x 5 x 5

Here, two 2s and two 5s are left which are not in a triplet. To make 100 a cube, we require one more 2 and one more 5.

Then, we obtain

100 x 2 x 5 = 2 x 2 x 2 x 5 x 5 x 5 = 1000 is a perfect cube

Hence, the smallest natural number by which 100 should be multiplied to make it a perfect cube is 2 x 5 = 10.

Answer needs Correction? Click Here

Q3 :  

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


Answer :

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Q4 :  

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?


Answer :

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Exercise 7.2 : Solutions of Questions on Page Number : 116

Q1 :  

Find the cube root of each of the following numbers by prime factorisation method.

(i) 64

(ii) 512

(iii) 10648

(iv) 27000

(v) 15625

(vi) 13824

(vii) 110592

(viii) 46656

(ix) 175616

(x) 91125


Answer :

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Q2 :  

State true or false.

(i) Cube of any odd number is even.

(ii) A perfect cube does not end with two zeroes.

(iii) If square 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.


Answer :

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Q3 :  

You are told that 1331 is a perfect cube. Can you guess without factorization what is its cube root? Similarly, guess the cube roots of 4913, 12167, 32768


Answer :

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Maths - NCERT Solution - Maths : CBSE NCERT Exercise Solutions for Class 8th for Cubes and Cube Roots ( Exercise 7.1, 7.2 ) will be available online in PDF book form soon. The solutions are absolutely Free. Soon you will be able to download the solutions.