# NCERT Solutions For Class 8 Maths Chapter 7

## NCERT Solutions Class 8 Maths Cubes and Cube Roots

NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots have been prepared by our team of subject experts with an objective to help students in their finals. These books include the solution for all the questions given in the NCERT maths textbooks. NCERT Solutions for Class 8 Maths Chapter 7 can be easily accessible by all the students either by downloading the PDF files or by studying online by visiting our website at BYJU’S. Those students who practicing questions from NCERT Solutions for Class 8 Maths Chapter 7 pdf, can gain more information about the chapter and can also have a quick review before their finals. The different types of NCERT solutions for class 8 maths teaches all the students about the various topics in the examination. Some of the different types of topics are Rational Numbers, Linear Equations in one variable and Practical Geometry.

### NCERT Solutions Class 8 Maths Chapter 7 Exercises

Exercise 7.1

Q1:

Mention the numbers that are not perfect cubes.

(A) 216

(B) 128

(C) 1000

(D) 100

(E) 46656

Solution:

(A) 216

Prime factors of 216: 2x2x2x3x3x3

Here all the factors are in the groups of 3’s

Therefore, 216 is said to be a perfect cube number.

 02 0216 02 0108 02 054 03 027 03 09 03 03 01

(B) 128

The prime factor of 128 = 2x2x2x2x2x2x2

Here one factor 2 does not appear in groups of 3

Hence, 128 is not a perfect cube.

 02 0128 02 064 02 032 02 016 02 08 02 04 02 02 01

(C) 1000

The prime factors of 1000 = 2x2x2x 5x5x5

Here all the factors are in groups of 3

Hence, 1000 is said to be a perfect cube.

 02 01000 02 0500 02 0250 05 0125 05 025 05 05 01

(D) 100

The prime factors of 100 is 2×2 x 5×5

Here all the factors do not appear in groups of 3.

Hence, 100 is not a perfect cube.

 02 0100 02 050 05 025 05 05 01

(E) 46656

The prime factors of 46656 = 2x2x2x2x2x2x 3 x3x3x3x3x 3

Here all the factors are in groups of 3

Hence, 46656 is said to be a perfect cube.

 02 046656 02 023328 02 011664 02 05832 02 02916 02 01458 03 0729 03 0243 03 081 03 027 03 09 03 03 01

Q2 :

Find the smallest number when multiplied to obtain a perfect cube:

(A) 243

(B) 256

(C) 72

(D) 675

(E) 100

Solution:

(A) 243

The prime factors of 243 = 3x3x3x3x 3

Here 3 does not appear in groups of 3

Hence, For 243 to be a perfect cube it should be multiplied by 3.

 03 0243 03 081 03 027 03 09 03 03 01

(B) 256

The prime factors of 256 is 2x2x2x2x2x 2 x2 x 2

Here one factor of 2 is required for it to make groups of 3.

Hence, for 256 to be a perfect cube it should be multiplied by 2.

 02 0256 02 0128 02 064 02 032 02 016 02 08 02 04 02 02 01

(C) 72

The prime factors for 72 = 2 x2x2x 3x 3

Here the factor 3 does not appear in groups of 3

Hence, For 72 to be a perfect cube it should be multiplied by 3.

(D) 675

The prime factors for 675 = 3x3x3x 5×5

Here the factor 5 does not appear in groups of 3

Hence, for 675 to be a perfect cube it should be multiplied by 5.

 03 0675 03 0225 03 075 05 025 05 05 01

(E) 100

The prime factors for 100 = 2x2x5x5

Here both the factors 2 and 5 are not in groups of 3

Hence, for 100 to be  a perfect cube it should be multiplied by 2 and 5. ( i.e. 2 x 5 =10 )

 02 0100 02 050 05 025 05 05 01

Q3:

Find the smallest number by which when divided obtain a perfect cube.

(A) 81

(B) 128

(C) 135

(D) 192

(E) 704

Solution:

(A) 81

The prime factors for 81 = 3 x 3 x 3 x 3

Here, there is one factor of 3 which extra from the group of 3

Hence, for 81 to be a perfect cube it should be divided by 3.

 03 081 03 027 03 09 03 03 01

(B) 128

The prime factors of 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2

Here there is one factor of 2 which in not in the group of 3

Hence, for 128 to be a perfect cube then it should be divided by 2.

 02 0128 02 064 02 032 02 016 02 08 02 04 02 02 01

(C) 135

The prime factors of 135 = 3 x 3 x 3 x 5

Here there is one factor of 5 which is not appearing with its group of 3.

Hence, for 135 to be a perfect cube it should be divided by 5.

 03 0135 03 045 03 15 05 05 01

(D)192

The prime factors for 192 = 2 x 2 x 2 x 2 x 2 x 2 x 3

Here there is one factor of 3 which does not appearing with its group of 3.

Hence for 192 to be a perfect cube then it should be divided by 3.

 02 0192 02 096 02 048 02 024 02 012 02 06 03 03 01

(E) 704

The prime factor for 704 = 2 x 2 x 2 x 2 x 2 x 2 x 11

Here there is one factor of 11 which is not appearing with its group of 3.

Hence for 704 to be a perfect cube it should be divided by 11.

 02 0704 02 0352 02 0176 02 088 02 044 02 022 02 011 01

Q4:

Reuben makes a cuboid of clay of sides 5 cm , 2 cm , 5 cm. If Reuben wants to form a cube how many such cuboids will be needed?

Solution:

The numbers given: 5 x 2 x 5

Since the factors of 2 and 4 are both not in groups of 3.

Then, the number should be multiplied by 2 x 2 x 5 = 20 for it to be made a perfect cube.

Hence Reuben needs 20 cuboids.

Exercise 7.2

Q1 :

By the method of prime factorization find the cube root for the following.

(A) 64

(B) 512

(C) 10648

(D) 27000

(E) 15625

(F) 13824

(G) 110592

(H) 46656

(I) 175616

(J) 91125

Solution:

(A) 64

643=2×2×2×2×2×23643=2×2$\sqrt[3]{64}=\sqrt[3]{2\times 2\times 2\times 2\times 2\times 2}\\\sqrt[3]{64}=2\times 2$

= 4

 02 064 02 032 02 016 02 08 02 04 02 02 01

(B) 512

5123=2×2×2×2×2×23$\sqrt[3]{512}=\sqrt[3]{2\times 2\times 2\times 2\times 2\times 2}$

= 2 x 2 x 2

= 8

 02 0512 02 0256 02 0128 02 064 02 032 02 016 02 08 02 04 02 02 01

(C) 10648

\sqrt[3]{10648}=\sqrt[3]{2\times 2\times 2\times 11\times 11\times 11}

= 2 x 11

=22

 02 010648 02 05324 02 02662 011 01331 011 0121 011 011 01

(D) 27000

270003=2×2×2×3×3×3×5×5×53$\sqrt[3]{27000}=\sqrt[3]{2\times 2\times 2\times 3\times 3\times 3\times 5\times 5\times 5}$

=>2 x 3 x 5

=>30

 02 027000 02 013500 02 06750 03 03375 03 01125 03 0375 05 0125 05 025 05 05 01

(E) 15625

156253=5×5×5×5×5×53$\sqrt[3]{15625}=\sqrt[3]{5\times 5\times 5\times 5\times 5\times 5}$

=> 5 x 5

=> 25

 05 015625 05 03125 05 0625 05 0125 05 025 05 05 01

(F) 13824

138243=2×2×2×2×2×2×2×2×2×3×3×33$\sqrt[3]{13824}=\sqrt[3]{2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 3\times 3\times 3}$

=> 2 x 2 x 2 x 3

=> 24

 02 13824 02 06912 02 03456 02 01728 02 0864 02 0432 02 0216 02 0108 02 054 03 27 03 09 03 03 01

(G) 110592

1105923=2×2×2×2×2×2×2×2×2×2×2×2×3×3×33$\sqrt[3]{110592}=\sqrt[3]{2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2 \times3\times 3\times 3}$

=> 2 x 2 x 2 x 2 x 3

=> 48

 02 0110592 02 055296 02 027648 02 013824 02 06912 02 03456 02 01728 02 0864 02 0432 02 0216 02 0108 02 054 03 027 03 09 03 03 01

(H) 46656

466563=2×2×2×2×2×2×3×3×3×3×3×33$\sqrt[3]{46656}=\sqrt[3]{2\times 2\times 2\times 2\times 2\times 2\times 3\times 3\times 3\times 3\times 3\times 3}$

=> 2 x 2 x 2 x 3 x 3 x 3

=> 36

(I) 175616

1756163=2×2×2×2×2×2×2×2×2×7×7×73$\sqrt[3]{175616}=\sqrt[3]{2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 7\times 7\times 7}$

=> 2 x 2 x 2 x 7

=> 56

 02 0175616 02 087808 02 043904 02 021952 02 010976 02 05488 02 02744 02 01372 02 0686 07 0343 07 049 07 07 01

(J) 91125

911253=3×3×3×3×3×3×5×5×53$\sqrt[3]{91125}=\sqrt[3]{3\times 3\times 3\times 3\times 3\times 3\times 5\times 5\times 5}$

=> 3 x 3 x 5

=> 45

 03 091125 03 010125 03 03375 03 01125 03 0375 05 0125 05 025 05 05 01

Q2:

State whether the following is true of false:

(A) Any off number of a cube is even.

(B) When a number end with two zeros, it is never a perfect cube.

(C) If the square of a given number ends with 5 then its cube will end with 25.

(D) There is no number that ends with 8 which is a perfect cube.

(E) The cube of a given two digit number will always be a three digit number.

(F) The cube of a two digit number will have either seven or more digits.

(G) The cube of single digit number may also be a single digit number.

Solution:

(A) The statement given is false.

Since, 13  = 1, 33 = 27, 53 = 125, . . . . . . . . . . are all odd.

(B) The given statement is true.

Since, a perfect cube ends with three zeroes.

Eg. 103 = 1000, 20= 8000, 303  = 27,000 , . . . . . . . . . . . . so on.

(C) The given statement is false

Since, 52 = 25, 53 = 125 , 152 = 225, 153 = 3375 ( Did not end with 25)

(D)  the given statement is false.

Since 123 = 1728 [the number ends with 8]

223 = 10648 [ the number ends with 8]

(E) The given statement is false

Since, 103 = 1000 [Four digit number]

And 113= 1331 [four digit number]

(F) The statement is False.

Since 993 = 970299 [Six digit number]

(G) the given statement is true

13 = 1 [single digit]

23 = 8 [single digit]

Q3 :

1331 is told to be a perfect cube. What are the factorization methods in which you can find its cube root? Similarly, find the cube roots for

(i)4913

(ii)12167

(iii)32768.

Solution:

We know that 103 = 1000 and possible cute of 113  = 1331

Since, the cube of units digit is 13  = 1

Then, cube root of 1331 is 11

(i) 4913

We know that 73 is 343

Next number that comes with 7 as the units place is 173  = 4913

Therefore the cube root of 4913 is 17

(ii) 12167

Since we know that 33 = 27

Here in cube, the ones digit is 7

Now the next number with 3 In the ones digit is 133 = 2197

And the next number with 3 in the ones digit is 233  = 12167

Hence the cube root of 12167 is 23

(iii) 32768

We know that 2= 8

Here in the cube, the ones digit is 8

Now the next number with 2 in the ones digit is 123  = 1728

And the next number with 2 as the ones digit 223 = 10648

And the next number with 2 as the ones digit 323 = 32768

Hence the cube root of 32768 is 32

Thus, these NCERT solutions are useful for training the students of class 8. These NCERT solutions for class 8 maths chapter 7 specifically deals with the different topics. Some of these different problems are perfect cubes, prime factorization etc. Thus, these are some of the various calculation metrics in chapter 8 . With the NCERT solutions for class 8  there are solutions for both maths and science. A student can download the PDFs below and go through the solutions for the problems. Thus these solutions are incredibly useful in providing the answers to the variety of questions asked. The solutions for math cubes and cube roots PDF can be downloaded easily and learnt whenever required. Thus, these are some of the NCERT solutions for class 8 maths chapter 7.