MCQ Questions for Class 8 Maths Cubes And Cube Roots Quiz 1

Cube of any odd number is even.

0%
True
0%
False
0%
Depends on the number
0%
Data Insufficient

Explanation

We know, cube of any odd number is odd.

Eg. The cube of the odd number

3

$3$ is

27

$27$ , which is an odd number.

Hence, the given statement is false.

Therefore, option

B

$B$ is correct.

Find the cube root of the following number by prime factorisation method:

64

$64$

0%
$2$
0%
$4$
0%
$8$
0%
$10$

Explanation

Prime factorising, we get,

64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2

$64 = 2\times2\times2\times2\times2\times2$

= 4 \times 4 \times 4 - -------- -

$= \underline {4\times4\times4}$ .

Here, the factor

4

$4$ occur as triplet. Hence, it is a perfect cube.

Therefore, cube root of

64

$64$ , i.e.

\sqrt[3]{64} = 4

$\sqrt[3]{64} = 4$ .

There is no perfect cube which ends with

8

$8$ .

0%
True
0%
False
0%
Insufficient Data
0%
None of these

Explanation

We know, cube of

2

$2$ , i.e.

2^{3} = 8

$2^3=8$ .

Here,

8

$8$ is a perfect cube.

That is, there is at least one perfect cube which ends with

8

$8$ .

Hence, the given statement is false.

Therefore, option

B

$B$ is correct.

Find the cube root of the following number by prime factorisation method:

512

$512$

0%
$6$
0%
$8$
0%
$7$
0%
$9$

Explanation

Prime factorising, we get,

$512 = \underline{2\times2\times2}\times\underline{2\times2\times2}\times\underline{2\times2\times2}$

$= \underline {8\times8\times8}$ .

Here, the factor $8$ occur as triplet. Hence, it is a perfect cube.

Therefore, cube root of $512$ , i.e. $\sqrt[3]{512} = 8$ .

\sqrt[3]{216}

$\sqrt[3]{216}$ is:

0%
Less than $6$
0%
Greater than $6$
0%
Equal to $6$
0%
Equal to $9$

Explanation

Prime factorising, we get,

$216= 6\times6\times6=6^3$ .

Then,

\sqrt[3]{216} = \sqrt[3]{6^{3}} = 6

$\sqrt[3]{216} = \sqrt[3] {6^3} = 6$ .

Hence, option

C

$C$ is correct.

8640

$8640$ is not a perfect cube.

0%
True
0%
False
0%
Ambiguous
0%
Data insufficient

Explanation

Prime factorization of:

8640 = 2^{6} \times 3^{3} \times 5^{1}

$8640=2^6 \times 3^3 \times 5^1$ .

We know, a perfect cube has multiples of

3

$3$ as powers of prime factors.

Here,

8640

$8640$ is not a perfect/complete cube.

That is, the given statement is true.

Hence, option

A

$A$ is correct.

Cubes of negative integers are negative.

0%
True
0%
False
0%
Ambiguous
0%
Data insufficient

Explanation

Multiplication of three negative numbers (i.e. the cube), will always be negative.

Eg.:

(- 4)^{3} = - 4 \times - 4 \times - 4 = 16 \times - 4 = - 64

$(-4)^3=-4\times -4 \times -4=16\times -4=-64$ , which is negative.

That is, the given statement is true.

Therefore, option

A

$A$ is correct.

Cube of all odd natural numbers are odd.

0%
True
0%
False
0%
Ambiguous
0%
Data insufficient

Explanation

We know, the multiplication of

3

$3$ odd numbers, i.e. the cube of an odd natural number, will always be odd.

For example, consider the odd natural numbers

3

$3$ and

5

$5$ .

Then, their cube is

3^{3} = 27

$3^3=27$ and

5^{3} = 125

$5^3=125$ , whose units place is odd.

That is, the cubes are also odd.

Hence, we can say, a cube of all odd natural numbers is odd.

Therefore, the given statement is true, and option

A

$A$ is correct.

Cubes of all even natural numbers are even.

0%
True
0%
False
0%
Ambiguous
0%
Data insufficient

Explanation

We know, the cube of an even natural number, will always be even

Example: Consider the even natural numbers

2

$2$ and

4

$4$ .

Then, their cubes are

2^{3} = 8

$2^3=8$ and

4^{3} = 64

$4^3=64$ , which are even numbers

Thus, we can show, cubes of all even natural numbers are even.

Therefore, the given statement is true and option

A

$A$ is correct.

What is the least number by which

8640

$8640$ is divided, the quotient as a complete cube number?

0%
$6$
0%
$7$
0%
$5$
0%
$8$

Explanation

$Prime\quad factorization\quad of\quad 8640,\\ 8640\quad =\quad 2^{ 6 }*3^{ 3 }*5\\ A\quad perfect\quad cube\quad has\quad multiples\quad of\quad 3\quad as\quad powers\quad of\quad prime\quad factors\\ i.e,\quad 8640\quad has\quad to\quad be\quad divided\quad by\quad 5\quad to\quad be\quad a\quad perfect\quad cube$

Calculate the value of

$\sqrt[3]{-64} + \sqrt{9^{2}}$ .

0%
$-4$
0%
$3$
0%
$5$
0%
$77$

Explanation

On prime factorising, we get,

$-64=(-4) \times (-4) \times (-4)$ $= (-4)^3$ .

Then,

$\sqrt[3]{-64} + \sqrt{9^{2}} = \sqrt[3]{(-4)^3} + 9 =-4+9= 5$ .

Hence, option

$C$ is correct.

Cube of

$(-3)$ is:

0%
$+27$
0%
$-9$
0%
$-6$
0%
$-27$

Explanation

Cube of

$(-3)$ is:

$(-3)^3=-3\times -3\times -3=-27$ .

Hence, option

$D$ is correct.

Cube root of the quotient of two negative integers is ______.

0%
positive
0%
zero
0%
negative
0%
none of these

Explanation

The quotient of two negative integers, i.e. a negative number when divided by another negative number gives a positive number.
Then, the cube root of the resultant positive number will be positive.

E.g.: Let the two integers be

$-16$ and

$-2$ .

Then, their quotient

$\dfrac{-16}{-2}=8$ , which is positive.

Therefore,

$\sqrt[3]{\dfrac{-16}{-2}}=\sqrt[3]{8}=2$ , which is also positive.

Hence, option

$A$ is correct.

The value of

$(3.1)^3$ is:

0%
$27.971$
0%
$29.791$
0%
$29.971$
0%
$27.197$

Explanation

Cube of

$(3.1)$ is:

$(3.1)^3=3.1 \times 3.1 \times 3.1$

$=29.791$ .

Hence, option

$B$ is correct.

A number raised to power

$3$ is called the _____.

0%
Cube of that number
0%
Square of that number
0%
Cube root of that number
0%
Square root of that number

Explanation

If a number is raised to the power

$3$ , then it is called the cube of that number.

Hence, option

$A$ is correct.

Cube of an odd natural number is an _____ number.

0%
odd
0%
even
0%
negative
0%
prime

Explanation

We know, the multiplication of odd natural numbers

$3$ times, i.e. the cube of an odd natural number, will always be odd.

That is because an odd number multiplied to another odd number, always yields an odd number.

For example, consider the odd natural numbers

$3$ and

$5$ .

Then, their cube is

$3^3=27$ and

$5^3=125$ , whose units place is odd.

That is, the cubes are also odd.

Hence, the cube of an odd natural number is an odd number.

Cube of even natural number is _____ number.

0%
even
0%
odd
0%
negative
0%
prime

Explanation

We know, the multiplication of

$3$ even numbers, i.e. the cube of an even natural number, will always be even

Example, consider the even natural numbers

$2$ and

$4$ .

Then, their cube is

$2^3=8$ and

$4^3=64$ , whose units place is even.

That is, the cubes are also even.

Hence, we can say, cube of even natural number is even.

Therefore, option

$A$ is correct.

The cube root of negative integer is _____.

0%
negative
0%
complex
0%
real
0%
positive

Explanation

Multiplication of

$3$ negative numbers (the cube of a negative number), will always be a negative number.

Therefore, the cube root of a negative integer is real and negative.

E.g.:

$\sqrt[3]{-8}=-2$ , which is real and negative.

Hence, options

$A$ and

$C$ are correct.

The cube of the given number is:

$0.4$ .

0%
$0.064$
0%
$0.074$
0%
$0.194$
0%
None

Explanation

Cube of the number

$0.4$ :

$(0.4)^{ 3 } = 0.4\times0.4\times 0.4 = 0.064$ .

Hence, option

$A$ is correct.

The cube of the given number is:

$1.3$ .

0%
$2.197$
0%
$2.187$
0%
$3.477$
0%
$8.447$

Explanation

Cube of the number

$1.3$ :

$(1.3)^{ 3 } = 1.3\times1.3\times1.3 = 2.197$ .

Hence, option

$A$ is correct.

The smallest natural number by which

$36$ must be multiplied to get a perfect cube is _____.

0%
$6$
0%
$216$
0%
$45$
0%
$2$

Explanation

Prime factorising $36$ , we get,

$36=2 \times 2 \times 3 \times 3=2^2 \times 3^2$ .

We know, a perfect cube has multiples of $3$ as powers of prime factors.

Here, number of $2$ 's is $2$ and number of $3$ 's is $2$ .

So we need to multiply another $2$ and $3$ in the factorization to make $36$ a perfect cube.

Hence, the smallest number by which $36$ must be multiplied to obtain a perfect cube is $2\times 3 =6$ .

Hence, option

$A$ is correct.

Choose the correct option:

There is no perfect cube which ends in

$4$ .

0%
True
0%
False
0%
Ambiguous
0%
Data insufficient

Explanation

We know,

cube of

$4$ , i.e.

$4^3=64$ , which is a perfect cube.

That is, there exists a perfect cube which ends in

$4$ .

Therefore, the given statement is false and option

$B$ is correct.

Value of

$\sqrt [3]{0.008}$ is:

0%
$0.2$
0%
$0.4$
0%
$0.002$
0%
None

Explanation

We know,

$0.008 = 8 \times { 10 }^{ -3 }=8 \times \dfrac {1}{10} \times \dfrac{1}{10} \times \dfrac {1}{10}$ .

Also, on prime factorising

$8$ , we get,

$8=2\times 2\times 2=2^3$ .

Then,

$\sqrt [ 3 ]{ 0.008 }$

$=\sqrt[3]{8 \times \dfrac {1}{10} \times \dfrac{1}{10} \times \dfrac {1}{10}}$

$=\sqrt[3]{2^3 \times \dfrac {1}{10} \times \dfrac{1}{10} \times \dfrac {1}{10}}$

$=2\times \dfrac {1}{10}=0.2$ .

Hence, option

$A$ is correct.

The cube of

$-3.1$ is:

0%
$-29.791$
0%
$-2.6891$
0%
$-2.5781$
0%
None

Explanation

Cube of the number

$-3.1$ :

$(-3.1)^{ 3 } = -3.1\times-3.1\times -3.1 = -29.791$ .

Hence, option

$A$ is correct.

Find the cube of

$0.6$ .

0%
$0.216$
0%
$0.36$
0%
$21.6$
0%
$2.16$

Explanation

Cube of

$0.6$ is:

$(0.6)^3=0.6\times 0.6\times 0.6$

$=0.216$ .

Hence, option

$A$ is correct.

Finding cube root is the inverse operation of finding the _____ .

0%
cube
0%
square
0%
reciprocal
0%
none of these

Explanation

Let

$y=x\times x\times x$ .
So,

$\sqrt[3] y =x$ .

Thus,

$x$ is a cube root of

$y$

and

$y$ is the cube of

$x$ .

Thus, we can say that finding cube root is the inverse of finding cube.

Hence, option

$A$ is correct.

Find the smallest number by which the following number must be multiplied to obtain a perfect cube:

$243$

0%
$3$
0%
$1$
0%
$0$
0%
$5$

Explanation

Prime factorizing

$243$ , we get,

$243=3\times 3\times 3\times 3\times 3=3^5$ .

We know, a perfect cube has multiples of

$3$ as powers of prime factors.
Here, number of

$3$ 's is

$5$ .

So we need to multiply another

$3$ in the factorization to make

$243$ a perfect cube.

Hence, the smallest number by which

$243$ must be multiplied to obtain a perfect cube is

$3$ .

Therefore, option

$A$ is correct.

Find

$\sqrt[3]{125}.$

0%
$10$
0%
$5$
0%
$-7$
0%
$15$

Explanation

On prime factorising, we get,

$125=5 \times 5\times 5$ $= 5^3$ .

Then, cube root of $125$ is:

$\sqrt[3]{125}=\sqrt[3]{5^3}= 5$ .

Therefore, option $B$ is correct.

Value of

$\displaystyle \sqrt[3]{343}$ is:

0%
$7$
0%
$-5$
0%
$\displaystyle \frac{7}{5}$
0%
$\displaystyle \frac{5}{7}$

Explanation

On prime factorising, we get,

${343}$

$= {7\times 7\times 7}$

$=7^3$ .

Then,

$\sqrt [3] {343}$

$=\sqrt [3] {7^3}$

$=7$ .

Hence, option

$A$ is correct.

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

192

0%
3
0%
4
0%
7
0%
1

Explanation

$Factorization\quad of\quad 192$

$192\quad =\quad 2\times2\times2\times2\times2\times2\times3$

$=\quad 2^{ 6 }\times3$

$To\quad make\quad a\quad perfect\quad cube,\quad we\quad need\quad to\quad have\quad multiples\quad of\quad 3\quad a\quad powers\quad of\quad prime\quad factors,\\ i.e,\quad we\quad divide\quad the\quad number\quad by\quad 3$

CBSE Questions for Class 8 Maths Cubes And Cube Roots Quiz 1 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

CBSE Questions for Class 8 Maths Cubes And Cube Roots Quiz 1 - MCQExams.com

0:0:2

Practice Class 8 Maths Quiz Questions and Answers

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