MCQ Questions for Class 8 Maths Cubes And Cube Roots Quiz 3

$(35)^3=$ ?

0%
$42875$
0%
$44875$
0%
$43005$
0%
None of these

Explanation

$\displaystyle 35=7\times 5$

$\displaystyle \therefore \quad { \left( 35 \right) }^{ 3 }={ \left( 7 \right) }^{ 3 }\times { \left( 5 \right) }^{ 3 }$

$\displaystyle =7\times 7\times 7\times 5\times 5\times 5$

$\displaystyle =343\times 125$

$\displaystyle =42875$

$\displaystyle \therefore$ The cube of

$35 = 42875.$

Find the unit digit of the cube root of the following number:

$13824$

0%
$8$
0%
$6$
0%
$4$
0%
$2$

Explanation

The unit's digit of

$13824$ is

$4.$
The only numbers whose cubes end with

$4$ are the numbers which have

$4$ in their unit's place. Hence cube root of

$13824$ has

$4$ in unit's place.

For an integer

$a,$ choose the correct statement.

0%
$\displaystyle { a }^{ 3 }$ is always greater than $\displaystyle { a }^{ 2 }$ .
0%
$\displaystyle { a }^{ 3 }$ is always smaller than $\displaystyle { a }^{ 2 }$ .
0%
$\displaystyle { a }^{ 2 }$ is always greater than $\displaystyle { a }^{ 3 }$ .
0%
None of the above

Explanation

False, since the square of 1 is 1 is equal to the cube of 1. In this case

$\displaystyle { a }^{ 3 }$ is not greater than

$a^2$ .

Find the unit digit of the cube root of the following number:

$226981$

0%
$3$
0%
$5$
0%
$7$
0%
$1$

Explanation

The unit's digit of

$226981$ is

$1.$
Since

$\displaystyle { \left( 1 \right) }^{ 3 }=1$ , the unit's digit of the cube root of

$226981$ is

$1.$

Find the unit digit of the cube root of the following number:

$571787$

0%
$5$
0%
$4$
0%
$3$
0%
$7$

Explanation

The unit's digit of

$571787$ is

$7.$
Since

$\displaystyle { \left( 3 \right) }^{ 3 }=27$ , the unit's digit of the cube root of

$571787$ is

$3.$

Find the unit digit of the cube root of the following number:

$175616$

0%
$5$
0%
$6$
0%
$8$
0%
$9$

Explanation

And the unit digit of cube root of

$6$ is

$6$

$\therefore$ The unit digit of the cube root of

$175616$ is

$6.$

What is the last digit (one's place) of

$\displaystyle { 143 }^{ 3 }?$

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

Explanation

Given, the number is $143^3$ .

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 $143$ , i.e. $143^3$ is $7$ .

Hence, option $C$ is correct.

What will be the unit digit of

$\displaystyle { 98765 }^{ 3 }$ .

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

Explanation

Given, the number is $98765^3$ .

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 , i.e. is $5$ .

Hence, option $A$ is correct.

Find the digit at unit's place of

$\displaystyle { 128 }^{ 3 }$ .

0%
$8$
0%
$6$
0%
$4$
0%
$2$

Explanation

Given, the number is $128^3$ .

Here, the units digit is $8$ .

We know, the cube of $8$ , i.e. $8^3=512$ , whose units place is $2$ .

Therefore, the units digit of the cube of $128$ , i.e. $128^3$ is $2$ .

Hence, option $D$ is correct.

$\displaystyle 27={ a }^{ 3 }$ , find the value of

$a$ .

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

Given,

$a^3=27$ .

Prime factorising, we get,

$27=3\times 3 \times 3$ .

Also,

$a=\sqrt[3]{a^3}$ .

Then, the value of

$a=\sqrt[3]{27}=\sqrt[3]{3^3}=3$

$\implies a=3$ .

Hence, option

$C$ is correct.

Cube of

$0.9$ is:

0%
$0.0729$
0%
$7.29$
0%
$729$
0%
$0.729$

Explanation

Cube of

$0.9$ :

$\displaystyle { 0.9 }^{ 3 }=0.9\times 0.9\times 0.9$

$\displaystyle =0.729$ .

Hence, option

$D$ is correct.

If the unit's digit of cube of a number is

$6$ , then what is the unit's digit of the number itself?

0%
$2$
0%
$4$
0%
$6$
0%
$8$

Explanation

Given, the units digit of cube of a number is $6$ .

Any number having $6$ in it's unit's place will have a $6$ in the unit's place of cube.It can be observed $...6\times ...6\times ...6=......6$

Therefore if the unit's digit of the cube is $6$ , the number must have $6$ in it's unit's place.

Hence, option $C$ is correct.

The ______ of a number is a number, whose multiplication by three times gives that number.

0%
cube root
0%
square root
0%
permutation
0%
repeating

Explanation

The cube root of a number is a number, when multiplication by three times gives that number.

Example:

$2 \times 2 \times 2 = 8.$

Here, the cube root of a number

$8$ , i.e.

$\sqrt[3]{8}=2$ is a number, when multiplication by three times (

$2\times2\times2$ ) gives that number, i.e.

$8$ .

Therefore, option

$A$ is correct.

What is the unit digit of

$\displaystyle { 100001 }^{ 3 }$ ?

0%
$1$
0%
$0$
0%
$3$
0%
$4$

Explanation

Given, the number is $100001^3$ .

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 $100001$ , i.e. is $1$ .

Hence, option $A$ is correct.

If a number is multiplied three times, then the number is called _____ of the final number.

0%
square root
0%
cube root
0%
absolute value
0%
negative

Explanation

If a number is multiplied three times, then the number multiplied is called cube root of the number.

Example: Consider the number

$2$ .
Then,

$2 \times 2 \times 2 = 8$ .
Thus, the number

$2$ is multiplied

$3$ times to give cube of the number

$2$ .

Hence, option

$B$ is correct.

$1728$ is an example of _____ of a number

$12$ .

0%
square
0%
cube
0%
Ramanujan number
0%
none of these

Explanation

Prime factorising

$1728$ , we get,

$1728$

$=(2\times2\times2)\times(2\times2\times2)\times(3\times3\times3)\\=2^3\times 2^3\times 3^3\\=12^3.$

$\therefore$

$5832$ is the cube of number

$12$ .

Hence, option

$B$ is correct.

Which odd number needs to be taken out to form a cube number?

$5 + 7 + 9 + 11$

0%
$7$
0%
$5$
0%
$9$
0%
$11$

Explanation

We know,

$\displaystyle 1=1={ 1 }^{ 3 }$ ,

$\displaystyle 3+5=8={ 2 }^{ 3 }$ ,

$\displaystyle 7+9+11=27={ 3 }^{ 3 }$ .

Therefore,

$5$ must be taken out to form a cube number,

$3^3=27$ .

Hence, option

$B$ is correct.

How will you represent

$49$ as a cube root?

0%
$\sqrt[7]{49}$
0%
$\sqrt[2]{49}$
0%
$\sqrt[4]{49}$
0%
$\sqrt[3]{49}$

Explanation

We know, the cube root is represented by the symbol

$\sqrt[3]{}.$
Then, the cube root of

$49$ is written as

$\sqrt[3]{49}.$

Hence, option

$D$ is correct.

$5832$ is an example of _____ of the number

$18.$

0%
square
0%
cube
0%
cube root
0%
square root

Explanation

Prime factorising

$5832$ , we get,

$5832$

$=(2\times2\times2)\times(3\times3\times3)\times(3\times3\times3)\\=2^3\times 3^3\times 3^3\\=18^3.$

$\therefore$

$5832$ is the cube of number

$18$ .

Hence, option

$B$ is correct.

To cube a number, how many times you need to multiply the number with itself?

0%
$1$ time
0%
$2$ times
0%
$3$ times
0%
$4$ times

Explanation

Consider a number

$a$ has

$b$ as its cube.
Then,

$b=a\times a\times a$
So, we need to multiply a number

$3$ times for it to be a cube.

Hence, option

$C$ is correct.

The cube root of a number is a number when ______ three times gives that number.

0%
divided
0%
added
0%
subtracted
0%
multiplied

Explanation

The cube root of a number is a number when multiplied three times gives that number.
Example:

$4 \times 4 \times 4 = 64.$

Here,

$4$ is the cube root of

$64.$

$\sqrt[3]{4\dfrac{12}{125}} =$ ?

0%
$1\dfrac{2}{5}$
0%
$1\dfrac{3}{5}$
0%
$1\dfrac{4}{5}$
0%
$2\dfrac{2}{5}$

Explanation

Simplifying,

$\sqrt[3]{4\dfrac{12}{125}}=\sqrt[3]{\dfrac{512}{125}}$ .

On prime factorisation of the numbers individually, we get,

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

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

$125=\underline { 5 \times 5 \times 5 }$ $=5^3$ .

Therefore,

$\sqrt[3]{4\dfrac{12}{125}}=\sqrt[3]{\dfrac{512}{125}}$

$=\sqrt[3]{\dfrac{8^3}{5^3}}={\dfrac{8}{5}}$ $={1\dfrac{3}{5}}$ .

Hence, option $B$ is correct.

Find the smallest number by which the number

$108$ must be multiplied to obtain a perfect cube.

0%
$2$
0%
$3$
0%
$4$
0%
$5$

Explanation

Prime factorising

$108$ , we get,

$108=3\times 3\times 3\times 2\times 2$

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

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

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

So we need to multiply another $2$ to the factorization to make $108$ a perfect cube.

Hence, the smallest number by which $108$ must be multiplied to obtain a perfect cube is $2$ .

Hence, option $A$ is correct.

From which of the following options, the perfect cube cannot end with?

0%
Two Zeroes
0%
Three Zeroes
0%
Six Zeroes
0%
None of these

Explanation

We know, the number of zeroes at the end of a number must be in multiples of three for it to be a perfect cube.

Here, only option

$A$ has number of zeros not equal to the multiple of

$3$ .

Hence, option

$A$ is correct.

Simplify:

$\displaystyle \sqrt[3]{\frac{1}{8}\times \frac{125}{64}}$ .

0%
$\displaystyle \frac{5}{8}$
0%
$\displaystyle \frac{375}{512}$
0%
$\displaystyle 2\frac{1}{2}$
0%
$\displaystyle 15\frac{5}{8}$

Explanation

On prime factorisation of the numbers individually, we get,

$8=\underline { 2 \times 2 \times 2 }$

$125=\underline { 5 \times 5 \times 5 }$

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

$64=\underline { 2 \times 2 \times 2 } \times \underline { 2 \times 2 \times 2 }$

$\Rightarrow \sqrt [ 3 ]{ 64 } =2 \times 2=4$ .

Then,

$\sqrt[3]{\cfrac{1}{8}\times \cfrac{125}{64}}=\sqrt[3]{\cfrac{1}{8}}\times \sqrt[3]{\cfrac{125}{64}}=\cfrac{\sqrt[3]{1}}{\sqrt[3]{8}}\times \cfrac{\sqrt[3]{125}}{\sqrt[3]{64}}$

$=\cfrac{1}{2}\times \cfrac{5}{4}=\cfrac{5}{8}$ .

Hence, option

$A$ is correct.

The smallest number by which

$8788$ must be divided so that the quotient is a perfect cube is _______.

0%
$4$
0%
$12$
0%
$16$
0%
$32$

Explanation

$\textbf{Step 1: Find the required number.}$

$\text{Prime factorising $$8788$$, we get,}$

$8788 = 2 \times 2 \times 13 \times 13 \times 13$

$= 2 ^ 2 \times 13 ^ 3$

$\text{We know, a perfect cube has prime factors as a group of $$3$$ .}$

$\text{Here, number of $$2$$'s is $$2$$ and number of $$13$$'s is $$3$$.}$

$\text{So we need to divide $$8788$$ by } 2^2 \text{ to make it a perfect cube.}$

$\text{Hence, the smallest number by which $$8788$$ must be divided to obtain a perfect cube is}$

$2^2=4$

$\textbf{Hence, option $$A$$ is correct.}$

$\displaystyle \sqrt[3]{333+\sqrt[3]{987+\sqrt[3]{2197}}}$ is equal to:

0%
$21$
0%
$18$
0%
$7$
0%
$3$

Explanation

$\sqrt [ 3 ]{ 333+\sqrt [ 3 ]{ 987+\sqrt [ 3 ]{ 2197 } } }$

$=\sqrt [ 3 ]{ 333+\sqrt [ 3 ]{ 987+\sqrt [ 3 ]{ 13\times 13\times 13 } } }$ ...[Since,

$2197=13 \times 13 \times 13$ ]

$=\sqrt [ 3 ]{ 333+\sqrt [ 3 ]{ 987+13 } }$

$=\sqrt [ 3 ]{ 333+\sqrt [ 3 ]{ 1000 } }$

$=\sqrt [ 3 ]{ 333+\sqrt [ 3 ]{ 10 \times 10\times 10 } }$ ...[Since,

$1000=10 \times 10 \times 10$ ]

$=\sqrt [ 3 ]{ 333+10 }$

$=\sqrt [ 3 ]{ 343 }$

$=\sqrt [ 3 ]{ 7\times 7\times 7 }$ ...[Since,

$343=7 \times 7\times 7$ ]

$=7$ .

Hence, option

$C$ is correct.

Cube of

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

0%
$-125$
0%
$\displaystyle-\frac{1}{125}$
0%
$\displaystyle \frac{1}{25}$
0%
$\displaystyle \frac{1}{125}$

Explanation

Cube of

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

$\displaystyle \left(\frac{1}{5}\right)^3$

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

Hence, option

$D$ is correct.

Sum of the digits of the smallest number by which

$1440$ should be multiplied so that it becomes a perfect cube is _____.

0%
$4$
0%
$6$
0%
$7$
0%
$8$

Explanation

Prime factorising $1440$ , we get,

$1440 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 5$

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

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

Here, number of $2$ 's is $5$ , number of $3$ 's is $2$ and number of $5$ 's is $1$ .

So we need to multiply another $2$ , $3$ and $5^2$ in the factorization to make $1440$ a perfect cube.

Hence, the smallest number by which $1440$ must be multiplied to obtain a perfect cube is $2\times 3 \times 5^2=150$ .

$\therefore$ The sum of digits of the smallest number is $= 1+5+0 = 6$ .

Hence, option $B$ is correct.

Find the smallest number by which

$9000$ should be divided so that the quotient becomes a perfect cube?

0%
$3$
0%
$9$
0%
$27$
0%
$1$

Explanation

Prime factorising $9000$ , we get,

$9000 = 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 5 \times 5$

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

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

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

So we need to remove $3^2$ from the factorization to make $9000$ a perfect cube.

Hence, the smallest number by which $9000$ must be divided to obtain a perfect cube is $3^2=9$ .

Therefore, option

$B$ is correct.

CBSE Questions for Class 8 Maths Cubes And Cube Roots Quiz 3 - 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 3 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

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