MCQ Questions for Class 8 Maths Cubes And Cube Roots Quiz 2

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

704

0%
11
0%
12
0%
14
0%
15

Explanation

$Factorization\quad of\quad 704$

$704\quad =\quad 2\times2\times2\times2\times2\times2\times11$

$=\quad 2^{ 6 }\times11$

$To\quad make\quad a\quad perfect\quad cube,\quad we\quad need\quad to\quad have\quad multiples\quad of\quad 3\quad a\quad powers\quad of\quad prime\quad factors,\\ i.e,\quad we\quad divide\quad the\quad number\quad by\quad 11$

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

135

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

Explanation

$Factorization\quad of\quad 135$

$135\quad =\quad 3\times3\times3\times5$

$=\quad 3^{ 3 }\times5$

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

0%
3
0%
6
0%
7
0%
14

Explanation

Writing 81 as a product of prime factors, we have

$81=\underline {3\times 3\times 3}\times 3$
Clearly, to make it a perfect cube, it must be divided by 3.

Cube of

$(-2)$ is _______.

0%
$+8$
0%
$-8$
0%
$-4$
0%
$+4$

Explanation

Cube of $(-2)$ is:

$(-2)^3=(-2)\times (-2)\times (-2)$

$=-8$ .

Hence, option $B$ is correct.

An odd cube number will have a/an ___ cube root.

0%
odd
0%
even
0%
fraction
0%
none of these

Explanation

We know, a cube of an even number (

$\text{applicable to any multiple of an even number}$ ) will always be even.
Then, cube root of an odd number can never be even.

This rules out option

$B$ .

We also know, there are perfect cubes which are odd numbers (

${27, 125, 343}$ , etc.)
Therefore, not all odd cube number need to be a fraction.

This rules out option

$C$ .

We know, when

$2$ or more odd numbers are multiplied, the resulting number will always be odd.
Then, any cube of an odd number will also given an odd number.
Therefore, any odd perfect cube will always have a

$\text{odd}$ cube root.

Hence, option

$A$ is correct.

The number

$x$ which when multiplied by itself three times gives a cube number

$y$ . Then,

$x$ is a _____ of

$y$ .

0%
cube root
0%
square root
0%
cube
0%
none of these

Explanation

Given,

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

$\sqrt[3] y =x$ .

Thus,

$x$ is a cube root of

$y$ .

Hence, option

$A$ is correct.

Cube of

$\displaystyle \left ( \frac{1}{5} \right )$ is

0%
-125
0%
$\displaystyle -\frac{1}{125}$
0%
$\displaystyle -\frac{1}{5}$
0%
$\displaystyle +\frac{1}{125}$

Explanation

$\displaystyle \frac{1}{5}\times \frac{1}{5}\times \frac{1}{5}=\frac{1}{125}$

An even cube number will have _____ cube root.

0%
an even
0%
an odd
0%
a fractional
0%
None of the above

Explanation

We know, even numbers end with:

$0,2,4,6,8$ .
Also, cubes of these even numbers will end with:

$0,8,64,216,512$ .

$\therefore$ They end in

$0,8,4,6,2$ respectively, which makes the cubes of an even number, an even number.
Thus even cube will have an even cube root.

Hence, option

$A$ is correct.

Find the value of:

$\displaystyle \sqrt[3]{3375}$ .

0%
$25$
0%
$35$
0%
$15$
0%
None of these

Explanation

On prime factorising, we get,

$3375=\underline{3\times 3\times 3}\times \underline{5\times 5\times 5}$

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

Then, value of $\sqrt[3]{3375}$ is:

$\sqrt[3]{3375}=\sqrt[3]{3^3 \times 5^3}= 3\times5=15$ .

Therefore, option $C$ is correct.

Find the value of:

$\displaystyle \sqrt[3]{8000}$ .

0%
$10$
0%
$40$
0%
$20$
0%
None of these

Explanation

On prime factorising, we get,

$8000=\underline{2\times 2\times 2}\times \underline{2\times 2\times 2}\times \underline{5\times 5\times 5}$

$=2^3\times 2^3\times 5^3$ .

Then, value of $\sqrt[3]{8000}$ is:

$\sqrt[3]{8000}=\sqrt[3]{2^3 \times 2^3 \times 5^3}= 2\times2\times5=20$ .

Therefore, option $C$ is correct.

Find the value of:

$\displaystyle \sqrt[3]{-216}$ .

0%
$-6$
0%
$-36$
0%
$32$
0%
$14$

Explanation

On prime factorising, we get,

$216=\underline{3\times 3\times 3}\times \underline{2\times 2\times 2}$

$=3^3\times 2^3=6^3$ .

Then, $-216$ $=(-6)^3$ .

Therefore, cube root of ${-216}$ ,

i.e. $\sqrt[3]{-216}=\sqrt[3]{(-6)^3}= -6$ .

Therefore, option $A$ is correct.

Find the value of:

$\displaystyle \sqrt[3]{9261}$ .

0%
$31$
0%
$21$
0%
$29$
0%
$19$

Explanation

On prime factorising, we get,

${9261}=\underline{3\times 3\times 3}\times \underline{7\times 7\times 7}=3^3 \times 7^3$ .

Then, value of $\sqrt[3]{9261}$ is:

$\sqrt[3]{9261}=\sqrt[3]{3^3\times 7^3}= 3 \times 7=21$ .

Therefore, option $B$ is correct.

Which symbol denotes cube root?

0%
$\sqrt [ 3 ]{ }$
0%
$\sqrt{ }$
0%
$\sqrt [ 5 ]{ }$
0%
$\sqrt [ 4 ]{ }$

Explanation

The symbol used to denote the cube root of a number is

$\displaystyle \sqrt [3]{}$ .

Hence, option

$A$ is correct.

Find the cube root of

$-3375$ .

0%
$-25$
0%
$-15$
0%
Not defined
0%
None of these

Explanation

On prime factorising, we get,

$3375=\underline{3\times 3\times 3}\times \underline {5\times5\times5}$ $=3^3 \times 5^3=15^3$ .

Then, $-3375$ $=(-15)^3$ .

Therefore, value of $\sqrt[3]{-3375}$ is:

$\sqrt[3]{-3375}=\sqrt[3]{(-15)^3}= -15$ .

Therefore, option $B$ is correct.

Write the units digit of the cube for $$71$$.71

0%
$1$
0%
$2$
0%
$5$
0%
$3$

Explanation

Given, the number is

$71$ .

Here, the units digit is

$1$ .

We know, the cube of

$1$ , i.e.

$1^3=1$ , whose units place is

$1$ .

Therefore, the units digit of the cube of

$71$ is

$1$ .

Hence, option

$A$ is correct.

Find the cube roots of :

$-1$ .

0%
$1$
0%
$-1$
0%
$\sqrt{-1}$
0%
Cannot be determined

Explanation

${\textbf{Step -1: Given, number is - 1.}}$

${\text{As we know that, prime factorization of 1 is,}}$

$1 = 1 \times 1 \times 1 = {1^3}$

${\text{So, prime factorization of - 1 will be,}}$

$- 1 = \left( { - 1} \right) \times \left( { - 1} \right) \times \left( { - 1} \right) = {\left( { - 1} \right)^3}$

$\Rightarrow - 1 = {\left( { - 1} \right)^3}$

${\textbf{Step -2: Taking cube root on both sides}}$

$\Rightarrow \sqrt[3]{{ - 1}} = \sqrt[3]{{{{\left( { - 1} \right)}^3}}} = - 1.$

${\text{Thus, cube root of - 1 is - 1}}{\text{.}}$

${\textbf{ Hence, Option (B)}}{\textbf{ -1, is correct answer.}}$

Find the cube root of:

$-216$ .

0%
$36$
0%
$-36$
0%
$6$
0%
$-6$

Explanation

On prime factorising, we get,

$216=\underline{3\times 3\times 3}\times \underline{2\times 2\times 2}$

$=3^3\times 2^3=6^3$ .

Then, $-216$ $=(-6)^3$ .

Therefore, cube root of ${-216}$ is:

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

Therefore, option $D$ is correct.

Find the smallest number by which

$8575$ must be multiplied so that the product is a perfect cube.

0%
$3$
0%
$7$
0%
$5$
0%
$9$

Explanation

Prime factorising $8575$ , we get,

$8575= 5 \times 5 \times 7 \times 7 \times 7$

$= 5^2 \times 7 ^3$ .

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

Here, number of $5$ 's is $2$ and number of $7$ 's is $3$ .

So we need to multiply another $7$ to the factorization to make $8575$ a perfect cube.

Hence, the smallest number by which $8575$ must be multiplied to obtain a perfect cube is $5$ .

Hence, option $C$ is correct.

$\displaystyle \sqrt[3]{-512}=$ ?

0%
$-64$
0%
$-8$
0%
Not defined
0%
None of these

Explanation

On prime factorising, we get,

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

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

Then, $-512$ $=(-8)^3$ .

Therefore, cube root of ${-512}$ ,

i.e. $\sqrt[3]{-512}=\sqrt[3]{(-8)^3}= -8$ .

Therefore, option $B$ is correct.

Write the units digit of the cube of

$4276$ .

0%
$6$
0%
$8$
0%
$4$
0%
none of these

Explanation

Given, the number is $4276$ .

Here, the units digit is $6$ .

We know, the cube of $6$ , i.e. $6^3=216$ , whose units place is $6$ .

Therefore, the units digit of the cube of $4276$ is $6$ .

Hence, option $A$ is correct.

Write the units digit of the cube of

$833$ .

0%
$3$
0%
$7$
0%
$1$
0%
$9$

Explanation

Given, the number is $833$ .

Here, the units digit is $3$ .

We know, the cube of $3$ , i.e. $3^3=27$ , whose units place is $7$ .

Therefore, the units digit of the cube of $833$ is $7$ .

Hence, option $B$ is correct.

No perfect cube can end with exactly:

0%
two zeros
0%
three zeros
0%
no zeros
0%
$7$

Explanation

A perfect cube of a number is obtained by multiplying the number by itself three times.

Eg:

${ 1 }^{ 3 } = 1\times1\times1= 1$

${ 2 }^{ 3 } =2\times2\times2 =8$

${ 3 }^{ 3 }=3\times3\times3 = 27$ .

Similarly,

${ 10 }^{ 3 } = 10 \times 10 \times 10 = 1000$ .

So, perfect cubes can never end with one or two zeroes and have to end with exactly

$3$ zeroes or number of zeros equal to the multiple of

$3$ .

Hence, option

$A$ is correct.

Write the units digit of the cube for

$125125125$ .

0%
$5$
0%
$7$
0%
$0$
0%
none of these

Explanation

Given, the number is $125125125$ .

Here, the units digit is $5$ .

We know, the cube of $5$ , i.e. $5^3=125$ , whose units place is $5$ .

Therefore, the units digit of the cube of $125125125$ is $5$ .

Hence, option $A$ is correct.

From the following options, choose the option with which perfect cube does not ends with:

0%
$5$
0%
$4$
0%
$0$
0%
None of the above

Explanation

We know,

$1^3 = 1$

$2^3 = 8$

$3^3 = 27$

$4^3 = 64$

$5^3 = 125$

$6^3 = 216$

$7^3 = 343$

$8^3 = 512$

$9^3 = 729$

$10^3 = 1000$ .

A number ending with

$1, 2, 3, 4, 5, 6, 7, 8, 9, 0$ has

perfect cubes ending with

$1, 8, 7, 4, 5, 6, 3, 2, 9, 0$ , respectively.

Hence, option

$D$ is correct.

Write the units digit for the cube of

$5922$ .

0%
$8$
0%
$4$
0%
$6$
0%
none of these

Explanation

Given, the number is $5922$ .

Here, the units digit is $2$ .

We know that, for the cube of $2$ , i.e. $2^3=8$ , the unit digit is $8$ .

Therefore, the units digit of the cube of $5922$ is $8$ .

Hence, option $A$ is correct.

$(56)^3=$ ?

0%
$175616$
0%
$235616$
0%
$175656$
0%
None of these

Explanation

$\displaystyle 56=7\times 8$

$\displaystyle \therefore \quad { \left( 56 \right) }^{ 3 }={ \left( 7 \right) }^{ 3 }\times { \left( 8 \right) }^{ 3 }$

$\displaystyle =7\times 7\times 7\times 8\times 8\times 8$

$\displaystyle =343\times 512$

$\displaystyle =175616$

$\displaystyle \therefore$ The cube of

$56 = 175616.$

$(402)^3=$ ?

0%
$64964808$
0%
$64964804$
0%
$645064802$
0%
None of these

Explanation

$\displaystyle 402=2\times 3\times 67$

$\displaystyle \therefore \quad { \left( 402 \right) }^{ 3 }={ \left( 2 \right) }^{ 3 }\times { \left( 3 \right) }^{ 3 }\times { \left( 67 \right) }^{ 3 }$

$\displaystyle =2\times 2\times 2\times 3\times 3\times 3\times 67\times 67\times 67$

$\displaystyle =8\times 27\times 300763$

$\displaystyle =64964808$

$\displaystyle \therefore$ The cube of

$402 = 64964808.$

$(72)^3=$ ?

0%
$373248$
0%
$473258$
0%
$383244$
0%
None of these

Explanation

$\displaystyle 72=8\times 9$

$\displaystyle \therefore \quad { \left( 72 \right) }^{ 3 }={ \left( 8 \right) }^{ 3 }\times { \left( 9 \right) }^{ 3 }$

$\displaystyle =8\times 8\times 8\times 9\times 9\times 9$

$\displaystyle =512\times 729$

$\displaystyle =373248$

$\displaystyle \therefore$ The cube of

$72 = 373248.$

Write the units digit of the cube for

$44447$ .

0%
$3$
0%
$7$
0%
$9$
0%
$1$

Explanation

Given, the number is $44447$ .

Here, the units digit is $7$ .

We know, the cube of $7$ , i.e. $7^3=343$ , whose units place is $3$ .

Therefore, the units digit of the cube of $44447$ is $3$ .

Hence, option $A$ is correct.

Cube root of

$343$ .

0%
$10$
0%
$9$
0%
$8$
0%
$7$

CBSE Questions for Class 8 Maths Cubes And Cube Roots Quiz 2 - 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 2 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

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