# Important Questions cube and cube roots class 8 chapter 7

Mathematics Chapter 7 of Class 8 is about Cubes and Cube Roots. The cube of any number is that number raised to the power of 3. When a number is multiplied three times by itself, we can say that the number has been cubed, and the product is called a cube of that number. If a is a number, then the cube of a is a ³ = a × a × a.

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## Get Access to CBSE Class 8 Maths Important Questions 2022-23 with Chapter-Wise Solutions

You can also find CBSE Class 8 Maths Chapter-by-Chapter Important Questions here:

### CBSE Class 8 Maths Important Questions

Sr No. Chapters Chapters Name
1 Chapter 1 Rational Numbers
2 Chapter 2 Linear Equations in One Variable
4 Chapter 4 Practical Geometry
5 Chapter 5 Data Handling
6 Chapter 6 Squares and Square Roots
7 Chapter 7 Cubes and Cube Roots
8 Chapter 8 Comparing Quantities
9 Chapter 9 Algebraic Expressions and Identities
10 Chapter 10 Visualising Solid Shapes
11 Chapter 11 Mensuration
12 Chapter 12 Exponents and Powers
13 Chapter 13 Direct and Inverse Proportions
14 Chapter 14 Factorisation
15 Chapter 15 Introduction to Graphs
16 Chapter 16 Playing with Numbers

## Important Questions Class 8 Mathematics Chapter 7 – With Solutions

Our in-house Mathematics faculty  has collated an entire list of Important Questions Class 8 Mathematics Chapter 7 by referring to various sources. For each and every question, the team has developed a step-by-step explanation that will help students understand the concepts used in each question. Also, the questions are chosen in a way that would cover the  complete chapter with intext and chapter end exercises. . So by practising from our question bank, students will be able to revise the chapter and boost their understanding of strong and weak points. And enhance their preparation by further focusing on weak areas which need more attention instead of trying to maximise their potential by cramming. Scoring a high percentage of marks is not difficult. You just require the right  strategy and correct understanding of the concepts.

Given below are a few of the questions and answers from our question bank of Class 8 Mathematics Chapter 7 Important Questions:

Question 1: A cuboid of plasticine made by Parikshit with sides 5 cm, 2 cm, and 5 cm. How many such cuboids will be needed to form a cube?

Answer 1: The given side of the cube is 5 cm, 2 cm and 5 cm.

Therefore, volume of cube = 5×2×5 = 50

The prime factorisation of 50 = 2×5×5

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

Therefore, we will multiply 50 by 2×2×5 = 20 to get the perfect square.

Hence, 20 cuboids are needed to form a cube.

Question 2: State true or false.

(i) The cube of any odd number is even

(ii) A perfect cube never ends with two zeros.

(iii) If the 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.

(i) This statement is false.

Taking a cube of any required odd numbers

3³= 3 x 3 x 3 = 27

7³=7 x 7 x 7= 343

5³=5 x 5 x 5=125

All the required cubes of any given odd number will always be odd.

(ii) This statement is true.

10³= 10 x 10 x 10= 1000

20³ = 20 x 20 x 20 = 2000

150³ =150 x150 x150 = 3375000

Hence a perfect cube will never end with two zeros.

(iii) This statement is false.

15²= 15 x15= 225

15³= 15 x 15 x 15= 3375

Thus, the square of any given number ends with 5; then the cube ends with the number 25 is an incorrect statement.

(iv) This statement is false.

2³= 2x2x2= 8

12³ = 12 x 12 x 12= 1728

Accordingly, There are perfect cubes ending with the number 8

(v) This statement is false.

The minimum two digits number is 10

And

10³=1000→4 Digit number.

The maximum two digits number is 99

And

99³=970299→6 Digit number

Accordingly, the cube of two-digit numbers can never be a three-digit number.

(vi) This statement is false

10³=1000→4 Digit number.

The maximum two digits number is 99

And

99³=970299→6 Digit number

Accordingly, the cube of two-digit numbers can never have seven or more digits.

(vii) This statement is true

1³ = 1 x 1 x 1= 1

2³ = 2 x 2 x 2= 8

According to the cube, a single-digit can be a single-digit number.

Question 3: Find the cube root of 91125 by the prime factorisation method.

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

Thus , 45 is the cube root of 91125.

Question 4: Find the cube of 3.5.

Answer 4: 3.53 = 3.5 x 3.5 x 3.5

= 12.25 x 3.5

= 42.875

Question 5: (1.2) ³ = _________.

Answer 5: (1.2) ³  = 12/10

= (12/10) × (12/10) × (12/10)

= 1728/1000

= 1.728

Question 6: There are _________ perfect cubes between 1 and 1000.

There are 8 perfect cubes between 1 and 1000.

2 × 2 × 2 = 8

3 × 3 × 3 = 27

4 × 4 × 4 = 64

5 × 5 × 5 = 125

6 × 6 × 6 = 216

7 × 7 × 7 = 343

8 × 8 × 8 = 512

9 × 9 × 9 = 729

Question 7: The cube of 100 will have _________ zeroes.

Answer 7: The cube of 100 will have  six zeroes.

= 1003

= 100 × 100 × 100

= 1000000

Question 8: Is 392 a perfect cube? If not, find the smallest natural number by which 392 should be multiplied so that the product is a perfect cube.

Answer 8: The prime factorisation of 392 gives:

392 = 2 x 2 x 2 x 7 x 7

As we can see, number 7 cannot be paired in a group of three. Therefore, 392 is not a perfect cube.

We must multiply the 7 by the original number to make it a perfect cube.

Thus,

2 x 2 x 2 x 7 x 7 x 7 = 2744, which is a perfect cube, such as 23 x 73 or 143.

Hence, the smallest natural number, which should be multiplied by 392 to make a perfect cube, is 7.

Question 9: Find the cube root of 10648 by the prime factorisation method.

Grouping the factors in triplets of number equal factors,

10648 = (2×2×2)×(11×11×11)

Here, 10648 can be grouped into triplets of number equal factors,

∴ 10648 = 2 ×11 = 22

Therefore, the cube root of 10648 is 22.

Question 10: Which of the following numbers are in perfect cubes? In the case of a perfect cube, find the number whose cube is the given number 256

Answer 10: A perfect cube can be expressed as a product of three numbers of equal factors

Resolving the given number into prime factors, we obtain

256 = 2 × 2 × 2 × 2 × 2× 2 × 2 × 2

Since the number 256 has more than three factors

∴ 256 is not a perfect cube.

Question 11: (13/10) ³

Answer 11: The cube of a rational number is the result of multiplying a number by itself three times.

To evaluate the cube of (13/10) ³

Firstly we need to convert into proper fractions, i.e.(13/10) ³

We need to multiply the given number three times, i.e. (13/10) × (13/10) × (13/10) =

(2197/1000)

∴ the cube of (1 3/10) is (2197/1000)

Question 12: Find the smallest number by which 128 must be divided to get a perfect cube.

Answer 12:  The prime factorisation of 128 is given by:

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 triples of equal factors.

Therefore, to obtain a perfect cube, we will divide 128 by 2.

Question 13: Find out the cube root of 13824 by the prime factorisation method.

Answer 13: First, let us prime factorise 13824:

13824 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3

= 2 ³ × 2 ³ × 2 ³ × 3 ³

3√13824 = 2 × 2 × 2 × 3 = 24

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Q.1 Show that 189 is not a perfect cube.

Marks:1
Ans

Prime factors of 189 = 3 — 3 — 3 — 7
Making triplets, one triplet is formed 3 — 3 — 3, leaving one more factor 7.

So, 189 cannot be expressed as a product of triplets.
Hence, 189 is not a perfect cube.

Q.2 Find the cube root of the following by prime factorisation method.
(i) 8000
(ii) 13824

Marks:2
Ans

$\begin{array}{l}\left(\mathrm{i}\right)\text{}\mathrm{Prime}\text{factorisation of 8000}\\ \\ 2\overline{)8000}\\ 2\overline{)4000}\\ 2\overline{)2000}\\ 2\overline{)1000}\\ 2\overline{)500}\\ 2\overline{)250}\\ 5\overline{)125}\\ 5\overline{)25}\\ \text{}5\\ \text{8000 =}\underset{¯}{\text{2}—\text{2}—\text{2}}—\underset{¯}{\text{2}—\text{2}—\text{2}}—\underset{¯}{5—5—5}\\ \\ \sqrt{8000}=\sqrt{\underset{¯}{\text{2}—\text{2}—\text{2}}—\underset{¯}{\text{2}—\text{2}—\text{2}}—\underset{¯}{5—5—5}}\\ \text{= 2}—\text{2}—\text{5}\\ \text{= 20}\\ \text{Thus, the cube root of 8000 is 20.}\\ \\ \left(\mathrm{ii}\right)\text{}\mathrm{Prime}\text{factorisation of 13824}\\ \\ 2\overline{)13824}\\ 2\overline{)6912}\\ 2\overline{)3456}\\ 2\overline{)1728}\\ 2\overline{)864}\\ 2\overline{)432}\\ 2\overline{)216}\\ 2\overline{)108}\\ 2\overline{)54}\\ 3\overline{)27}\\ 3\overline{)9}\\ \text{}3\\ \text{}\\ \text{13824 =}\underset{¯}{\text{2}—\text{2}—\text{2}}—\underset{¯}{\text{2}—\text{2}—\text{2}}—\underset{¯}{2—2—2}—\underset{¯}{3—3—3}\\ \\ \sqrt{13824}=\sqrt{\underset{¯}{\text{2}—\text{2}—\text{2}}—\underset{¯}{\text{2}—\text{2}—\text{2}}—\underset{¯}{2—2—2}—\underset{¯}{3—3—3}}\\ \text{= 2}—\text{2}—2—3\\ \text{= 24}\\ \text{Thus, the cube root of 13824 is 24.}\end{array}$

Q.3 Find the cube root of 32768 through estimation.

Marks:1
Ans

The given number is 32768.
Form groups of three starting from the rightmost digit of 32768.
In this case it is 768 and 32 has only 2 digits.
Taking 768, 8 in unit digit, the required cube root is 2.
The other group i.e. 32, as cube of 3 is 27 and cube of 4 is 64, 32 lies between 27 and 64.
The smallest among 3 and 4 is 3.
The ones place of 3 is itself.
Take 3 as tens place of the cube root of 32768.
Thus, cube root of 32768 is 32.

Q.4 Find the smallest number by which 704 must be divided to obtain a perfect cube.

Marks:2
Ans

$\begin{array}{l}704=\underset{¯}{2—2—2}—\underset{¯}{2—2—2}—11\\ \mathrm{The}\text{}\mathrm{prime}\text{}\mathrm{factor}\text{}11\text{}\mathrm{does}\mathrm{not}\text{}\mathrm{appear}\text{}\mathrm{in}\text{}\mathrm{a}\text{}\mathrm{group}\text{}\mathrm{of}\text{}\mathrm{three}.\\ \mathrm{So},704\text{}\mathrm{is}\text{}\mathrm{not}\text{}\mathrm{a}\text{}\mathrm{perfect}\text{}\mathrm{cube}.\\ \mathrm{To}\text{}\mathrm{make}\text{}\mathrm{it}\text{}\mathrm{a}\text{}\mathrm{perfect}\text{€‰}\mathrm{cube},\text{}\mathrm{divide}\text{}\mathrm{it}\text{}\mathrm{by}\text{}11.\\ \mathrm{Thus},704·11=\underset{¯}{2—2—2}—\underset{¯}{2—2—2}\\ \text{}=64,\text{}\mathrm{which}\text{}\mathrm{is}\text{}\mathrm{a}\text{}\mathrm{perfect}\text{}\mathrm{cube}.\\ \mathrm{Hence},\mathrm{the}\mathrm{required}\mathrm{smallest}\mathrm{number}\mathrm{is}11.\end{array}$