MCQ Questions for Class 8 Maths Cubes And Cube Roots Quiz 6

The smallest number by which

$5400$ must be multiplied so that it becomes a perfect cube is:

0%
$12$
0%
$10$
0%
$5$
0%
$3$

Explanation

Prime factorising

$5400$ , we get,

$5400 = 3 \times 3 \times 3 \times 5 \times 5 \times 2 \times 2 \times 2$

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

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

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

So we need to multiply another $5$ to the factorization to make $5400$ a perfect cube.

Hence, the smallest number by which $5400$ must be multiplied to obtain a perfect cube is $5$ .

Therefore, option $C$ is correct.

Find the cube root of

$2744$ by estimation method.

0%
$12$
0%
$14$
0%
$24$
0%
$16$

Explanation

Step

$I$ : Form groups of

$3$ starting from right most digit of

$2744$ , i.e.

$\underline{2}\underline{744}$ .
Then, the two groups are

$744$ and

$2.$

Here,

$744$ has

$3$ digit and

$2$ has only

$1$ .

Step

$II$ : Take

$744$ .
Digit in unit place

$= 4$ .
Therefore, we take one's place of required cube root as

$4$ ....[Since,

$4^3=64$ ].

Step

$III$ : Now, take the other group

$2$ .
We know,

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

${ 2 }^{ 3 }=8$ .
Here, the smallest number among

$1$ and

$2$ is

$1$ .
Therefore, we take

$1$ as ten's place.

$\displaystyle \therefore \quad \sqrt[3] { 2744 } =14$ .

Hence, option

$B$ is correct.

Cube root of the product of two negative numbers is ____.

0%
positive
0%
negative
0%
positive or negative
0%
none of these

Explanation

The product of two negative integers, i.e. a negative number when multiplied 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

$-4$ and

$-2$ .

Then, their quotient

$(-4)\times(-2)=8$ , which is positive.

Therefore,

$\sqrt[3]{(-4)\times(-2)}=\sqrt[3]{8}=2$ , which is also positive.

Hence, option

$A$ is correct.

Which number should be multiplied by

$72$ to make it a perfect cube?

0%
$3$
0%
$2$
0%
$9$
0%
$8$

Explanation

Expressing

$72$ in terms of its factors, we get

$\Rightarrow 72 = 8 \times 9 = 2^3\times 3^2$

To make this a perfect cube, we need to multiply

$72$ by

$3$ which will give us

$2^3X3^3$ .

Find the smallest number by which

$500$ should be multiplied to make it a perfect cube.

0%
$5$
0%
$10$
0%
$3$
0%
$2$

Explanation

Prime factorising $500$ , we get,

$500 = 5 \times 5 \times 5 \times 2 \times 2$

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

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

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

So we need to multiply another $2$ to the factorization to make $500$ a perfect cube.

Hence, the smallest number by which $500$ must be multiplied to obtain a perfect cube is $2$ .

Therefore, option

$D$ is correct.

Cube root of

$0.000216$ is:

0%
$0.6$
0%
$0.06$
0%
$0.006$
0%
None of the above

Explanation

We know,

$0.000216=\dfrac{216}{1000000}$ .

Prime factorising, we get,

$216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3=6^3$

$1000000 = (2\times2 \times 2) \times(2\times2\times2)\times (5\times5\times5)\times(5\times5\times5)=2^3\times2^3\times5^3\times5^3=100^3$ .

Then, cube root of

$0.000216=\dfrac{216}{1000000}$ is:

$\sqrt[3]{0.000216}=\sqrt[3]{\dfrac{216}{1000000}}$

$=\sqrt[3]{\dfrac{6^3}{100^3}}={\dfrac{6}{100}}=0.06$ .

Hence, option

$B$ is correct.

How many consecutive odd numbers are needed to obtain sum as

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

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

Explanation

We know,

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

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

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

$\displaystyle \therefore$

$3$ consecutive odd numbers are required to obtain the sum as

$3^3$ .

Hence, option

$C$ is correct.

Find the cube root of

$166375$ by prime factorization method.

0%
$45$
0%
$55$
0%
$65$
0%
$75$

Explanation

On prime factorising, we get,

$166375= 5\times5\times5\times11\times11\times11$

$= 5^3\times11^3$ .

Then, cube root of $166375$ is:

$\sqrt[3]{166375}=\sqrt[3]{5^3 \times 11^3}=5\times11=55$ .

Therefore, option $B$ is correct.

Find the cube root of

$39304$ by estimation method.

0%
$24$
0%
$44$
0%
$34$
0%
$54$

Explanation

Step

$I$ : Form groups of

$3$ starting from right most digit of

$39304$ , i.e.

$\displaystyle \overline { 39 }\: \overline { 304 }$ .
Then, the two groups are

$39$ and

$304.$

Here,

$39$ has

$2$ digit and

$304$ has

$3$ digit.

Step

$II$ : Take

$304$ .
Digit in unit place

$= 4$ .
Therefore, we take one's place of required cube root as

$4$ ....[Since,

$4^3=64$ ].

Step

$III$ : Now, take the other group

$39$ .
We know,

$\displaystyle {3}^{ 3 }=27$ and

${4 }^{ 3 }=64$ .
Here, the smallest number among

$3$ and

$4$ is

$3$ .
Therefore, we take

$3$ as ten's place.

$\displaystyle \therefore \quad \sqrt[3] { 39304 } =34$ .

Hence, option

$C$ is correct.

The cube root of any multiple of

$8$ is always divisible by:

0%
$2$
0%
$4$
0%
$8$
0%
$16$

Explanation

Prime factorising

$8$ , we get,

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

$x$ is a cube and a multiple of

$8,$
then

$\displaystyle x={ 2 }^{ 3 }\times y$ .

So, cube root of

$x$ is divisible by

$2.$

Hence, option

$A$ is correct.

Cube of

$1.5$ is _________.

0%
$3.375$
0%
$33.75$
0%
$3.125$
0%
$31.25$

Explanation

Cube of

$1.5$ is:

$\displaystyle { \left( 1.5 \right) }^{ 3 }$

$={ \left( \dfrac { 15 }{ 10 } \right) }^{ 3 }$

$={ \left( \dfrac { 3 }{ 2 } \right) }^{ 3 }$

$=\cfrac { 27 }{ 8 }$

$=3.375$ .

Hence, option

$A$ is correct.

If the unit place of a number is

$7$ , then find the unit digit of its cube.

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

Explanation

Given, the units digit of the number 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 is $3$ .

Hence, option $B$ is correct.

Find the cube root of

$59319$ by estimation method.

0%
$29$
0%
$39$
0%
$49$
0%
$59$

Explanation

Step $I$ : Form groups of $3$ starting from right most digit of $59319$ , i.e. $\displaystyle \overline { 59 }\: \overline { 319 }$ .

Then, the two groups are $59$ and $319.$

Here, $59$ has $2$ digit and $319$ has $3$ digit.

Step $II$ : Take $319$ .

Digit in unit place $= 9$ .

Therefore, we take one's place of required cube root as $9$ ....[Since, $9^3=729$ ].

Step $III$ : Now, take the other group $59$ .

We know, $\displaystyle {3}^{ 3 }=27$ and ${4 }^{ 3 }=64$ .

Here, the smallest number among $3$ and $4$ is $3$ .

Therefore, we take $3$ as ten's place.

$\displaystyle \therefore \quad \sqrt[3] { 59319 } =39$ .

Hence, option $B$ is correct.

Find the smallest number which should be multiplied to

$392$ to make it a perfect cube.

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

Explanation

$\displaystyle 392={ 2 }^{ 3 }\times { 7 }^{ 2 }$
We require one more

$7$ for it to be a perfect cube.

What will be the unit digit of

$\displaystyle { 137959 }^{ 3 }$ .

0%
$1$
0%
$3$
0%
$6$
0%
$9$

Explanation

Given, the number is $137959^3$ .

Here, the units digit is $9$ .

We know, the cube of $9$ , i.e. $9^3=729$ , whose units place is $9$ .

Therefore, the units digit of the cube of , i.e. is $9$ .

Hence, option $D$ is correct.

If the unit digit of

$\displaystyle { x }^{ 3 }$ is

$3$ , then the unit digit of

$x$ is:

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

Explanation

For the options given, if unit digit of

$x$ is:

$\circ$

$3$ , then unit digit of

$x^3$ will be

$7$ (

$\because 3^3=27$ )

$\circ$

$5$ , then unit digit of

$x^3$ will be

$5$ (

$\because 5^3=125$ )

$\circ$

$7$ , then unit digit of

$x^3$ will be

$3$ (

$\because 7^3=343$ )

$\circ$

$1$ , then unit digit of

$x^3$ will be

$1$ (

$\because 1^3=1$ )

Hence,

$Op-C$ is correct.

Find the smallest number by which

$4232$ must be multiplied to make it a perfect cube.

0%
$2$
0%
$17$
0%
$19$
0%
$23$

Explanation

Prime factorising of $4232$ is as below,

$\displaystyle 4232=2\times 2\times 2\times 23\times 23$

$= 2 ^3 \times 23 ^2$ .

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

Here, number of $2$ 's is $3$ and number of $23$ 's is $2$ .

So we need to multiply another $23$ to the factorization to make $4232$ a perfect cube.

Hence, the smallest number by which $4232$ must be multiplied to obtain a perfect cube is $23$ .

Hence, option $D$ is correct.

$x={ 7 }^{ 2 }\times { 13 }^{ 1 }\times { 19 }^{ 3 }$ . Which smallest number should be multiplied by

$x$ to make it a perfect cube?

0%
$7$
0%
$13$
0%
$91$
0%
$1183$

Explanation

We can see

$x$ is expressed in terms of its prime factors
To make

$x$ a perfect cube we need to ensure the prime factors are raised to an exponent value which is a multiple of

$3$ .
In this case

$x = 7^2 \times 13^1 \times 19^3$ .

Here, number of $7$ 's is $2$ , number of $13$ 's is $1$ and number of $19$ 's is $3$ .

So we need to multiply another $7$ and $13^2$ to the factorization to make $x$ a perfect cube.

Hence, the smallest number by which $x$ must be multiplied to obtain a perfect cube is $7\times 13^2=1183$ .

Therefore, option $D$ is correct.

Find the cube root of

$3375$ by the method of prime factorization.

0%
$15$
0%
$25$
0%
$35$
0%
$55$

Explanation

On prime factorising, we get,

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

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

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

Therefore, option $B$ is correct.

If we write

$\displaystyle { n }^{ 3 }$ as the sum of consecutive odd numbers, then what will be the first term?

0%
$2n+1$
0%
$(2n+1)(2n-1)$
0%
$2n-1$
0%
$n(n-1)+1$

Explanation

From observation

$\displaystyle { 1 }^{ 3 }=1;1\times 0+1$ , where

$n=0$ .

$\displaystyle { 2 }^{ 3 }=3+5;3=2\times 1+1$ , where

$n=2$ .

$\displaystyle { 3 }^{ 3 }=7+9+11;7=3\times 2+1$ , where

$n=3$ .

Similarly, for

$\displaystyle { n }^{ 3 }$ , the first term will be

$n(n-1)+1$ .

Therefore, option

$D$ is correct.

Find the cube root of

$343000$ by estimation method.

0%
$60$
0%
$70$
0%
$80$
0%
$700$

Explanation

Step $I$ : Form groups of $3$ starting from right most digit of $343000$ , i.e. $\displaystyle \overline { 343 }\: \overline {000}$ .

Then, the two groups are $343$ and $000.$

Here, $343$ has $3$ digit and $000$ has $3$ digit.

Step $II$ : Take $000$ .

Digit in unit place $= 0$ .

Therefore, we take one's place of required cube root as $0$ .

Step $III$ : Now, take the other group $343$ .

We know, $\displaystyle {7}^{ 3 }=343$ .

Therefore, we take $7$ as ten's place.

$\displaystyle \therefore \quad \sqrt[3] { 343000 } =70$ .

Hence, option $B$ is correct.

What is the value of

$\displaystyle \sqrt [ 3 ]{ 125 } +\sqrt [ 3 ]{ 125 }$ ?

Use estimation method to find the cube root.

0%
$-10$
0%
$0$
0%
$5$
0%
$10$

Explanation

First, we need to find

$\sqrt[3]{125}$ with the estimation method.

STEP 1:

With the estimation method divide the number

$125$ into two parts. Since there are odd number of digits keep the heavier side on the right side.

Therefore it will be divided into only one part that is

$\underline{125}$ .

STEP 2:

Since the units place of the group is

$5$ , the cube root will also have a

$5$ in its units place....[Since,

$5^3=125$ ].

Therefore,

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

Then,

$\Rightarrow \sqrt[3]{125}+\sqrt[3]{125}=5+5=10$ .

Hence, option

$D$ is correct.

What is the value of

$\displaystyle \sqrt [ 3 ]{ 27 } \times \sqrt [ 3 ]{ -27 }$ ?

0%
$-9$
0%
$-27$
0%
$9$
0%
$0$

Explanation

On prime factorising, we get,

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

Then, $-27$ $=(-3)^3$ .

Therefore, value of $\sqrt[3]{27}\times\sqrt[3]{-27}$ is:

$\sqrt[3]{27}\times\sqrt[3]{-27}$

$=\sqrt[3]{3^3}\times\sqrt[3]{(-3)^3}$

$=3\times(-3)=-9$ .

Therefore, option $A$ is correct.

How many consecutive odd numbers are needed to make sum as

$\displaystyle { 13 }^{ 3 }$ ?

0%
$11$
0%
$13$
0%
$15$
0%
$17$

Explanation

We know,

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

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

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

$\displaystyle 13+15+17+19=64={ 4 }^{ 3 }$ ,

$\displaystyle 21+23+25+27+29=125={ 5 }^{ 3 }$ ,

$\displaystyle 31+33+35+37+39+41=216={ 6 }^{ 3 }$ ,

$\displaystyle 43+45+47+49+51+53+55=343={ 7 }^{ 3 }$ ,

$\displaystyle 57+59+61+63+65+67+69+71=512={ 8 }^{ 3 }$ ,

$\displaystyle 73+75+77+79+81+83+85+87+89=729={9 }^{ 3 }$ ,

$\displaystyle 91+93+95+97+99+101+103+105+107+109=1000={ 10 }^{ 3 }$ ,

$\displaystyle 111+113+115+117+119+121+123+125+127+129+131=1331={ 11 }^{ 3 }$ ,

$\displaystyle 133+135+137+139+141+143+145+147+149+151+153+155=1728={ 12 }^{ 3 }$ ,

$\displaystyle 157+159+161+163+165+167+169+171+173+175+177+179+181=2197={ 13 }^{ 3 }$ .

$\displaystyle \therefore$

$13$ consecutive odd numbers are required to obtain the sum as

$13^3$ .

Hence, option

$B$ is correct.

Find the smallest number which should be multiplied to

$231525$ to make it a perfect cube.

0%
$5$
0%
$3$
0%
$7$
0%
$21$

Explanation

Prime factorising $231525$ , we get,

$231525= 3 \times 3 \times 3 \times 5 \times 5 \times 7 \times 7 \times 7$

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

We know that a perfect cube number has its factor with an exponent as multiples of $3.$

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

So we need to multiply another $5$ to the factorization to make $2,31,525$ a perfect cube.

Hence, the smallest number by which $2,31,525$ must be multiplied to obtain a perfect cube is $5$ .

Hence, option $A$ is correct.

What is the value of cube root of

${ 4913 }$ ?

Use estimation method to find the cube root.

0%
$18$
0%
$15$
0%
$16$
0%
$17$

Explanation

Step $I$ : Form groups of $3$ starting from right most digit of $4913$ , i.e. $\displaystyle \overline { 4 }\: \overline { 913 }$

Then, the two groups are $4$ and $913.$

Here, $4$ has only $1$ digit and $913$ has $3$ digits.

Step $II$ : Take $913$

Digit in unit place $= 3$

Therefore, we take one's place of required cube root as $7$ .... [Since, $7^3=343$ ]

Step $III$ : Now, take the other group $4$

We search for the largest cube number which is less than the number in the second group.

Number in the second group is $4$

And also, we know, $1 \leq 4 < 8$

$\Rightarrow 1^3 \leq 4 < 2^3$

Here, the smallest number among $1$ and $2$ is $1$

Therefore, we take $1$ as ten's place.

$\displaystyle \therefore \quad \sqrt[3] { 4913 } =17$

Hence, option $D$ is correct.

Cube of

$1.3$ is:

0%
$2197$
0%
$219.7$
0%
$21.97$
0%
$2.197$

Explanation

$\displaystyle { 1.3 }^{ 3 }$

$=\dfrac { 13\times 13\times 13 }{ 10\times 10\times 10 }$

$=\dfrac { 2197 }{ 1000 }$

$=2.197$

Find the smallest number which should be multiplied to

$1352$ to get a perfect cube.

0%
$12$
0%
$13$
0%
$61$
0%
$21$

Explanation

By prime factorising $1352$ ,

we get, $1352 = 2 \times 2 \times 2 \times 13 \times 13$

$= 2 ^3 \times 13 ^2$

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

Here, number of $2$ 's is $3$ and number of $13$ 's is $2$ .

So we need to multiply another $13$ to the factorization to make $1352$ a perfect cube.

Hence, the smallest number by which $1352$ must be multiplied to obtain a perfect cube is $13$ .

What is the value of

$\displaystyle \sqrt [ 3 ]{ -8 } -\sqrt [ 3 ]{ -216 }$ ?

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

Explanation

On prime factorisation of the numbers individually, we get,

$216=\underline { 2 \times 2 \times 2 } \times \underline { 3 \times 3 \times 3 }=2^3\times3^3=6^3$ .

$\Rightarrow { -216 } =(-6)^3$ .

$8=\underline { 2 \times 2 \times 2 } =2^3$

$\Rightarrow { -8 } =(-2)^3$ .

Therefore,

$\sqrt[3]{-8}-\sqrt[3]{-216}$ $=\sqrt[3]{(-2)^3}-\sqrt[3]{(-6)^3}=(-2)-(-6)=-2+6=4$ .

Thus, option $C$ is correct.

What is the value of

$\displaystyle \sqrt [ 3 ]{ 64 } \div \sqrt [ 3 ]{ -64 }$ ?

0%
$-1$
0%
$0$
0%
$1$
0%
$16$

Explanation

On prime factorising, we get,

$64=\underline{4\times 4\times4}$ $=4^3$ .

Then, $-64$ $=(-4)^3$ .

Therefore, value of $\sqrt[3]{64}\div\sqrt[3]{-64}$ is:

$\sqrt[3]{64}\div\sqrt[3]{-64}$

$=\sqrt[3]{4^3}\div\sqrt[3]{(-4)^3}$

$=4\div(-4)=-1$ .

Therefore, option $A$ is correct.

CBSE Questions for Class 8 Maths Cubes And Cube Roots Quiz 6 - 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 6 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

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