MCQ Questions for Class 8 Maths Cubes And Cube Roots Quiz 7

Find the cube root of

$24389$ by estimation method.

0%
$27$
0%
$31$
0%
$29$
0%
$33$

Explanation

Step $I$ : Form groups of $3$ starting from right most digit of $24389$ , i.e. $\displaystyle \overline { 24 }\: \overline { 389 }$ .

Then, the two groups are $24$ and $389.$

Here, $24$ has $2$ digit and $389$ has $3$ digit.

Step $II$ : Take $389$ .

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 $24$ .

We know, $\displaystyle {3}^{ 3 }=27$ and ${2 }^{ 3 }=84$ .

Here, the smallest number among $3$ and $2$ is $2$ .

Therefore, we take $2$ as ten's place.

$\displaystyle \therefore \quad \sqrt[3] { 24389 } =29$ .

Hence, option $C$ is correct.

Which of the following number has same unit digit as its cube?

0%
$\displaystyle { 122 }^{ 3 }$
0%
$\displaystyle { 168 }^{ 3 }$
0%
$\displaystyle { 137 }^{ 3 }$
0%
$\displaystyle { 184 }^{ 3 }$

Explanation

Here,

$122^3$ :

The number ends in

$2$ and

$2^3=8$ so its cube will end in

$8$ .

$168^3$ :

The number ends in

$8$ and

$8^3=512$ so its cube will end in

$2$ .

$137^3$ :

The number ends in

$7$ and

$7^3=343$ so its cube will end in

$3$ .

$184^3$ :

The number ends in

$4$ and

$6^3=64$ so its cube will end in

$4$ .

Therefore, option

$D$ is correct.

Which odd number should be excluded to form a cube number?

$41 + 43 + 45 + 47 + 49 + 51 + 53 + 55$

0%
$43$
0%
$45$
0%
$47$
0%
$41$

Explanation

We know,

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

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

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

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

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

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

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

Therefore,

$41$ must be taken out to form a cube number,

$7^3=343$ .

Hence, option

$D$ is correct.

Express the sum of odd consecutive cube number for

$6^3$ .

0%
$31 + 33 + 35 + 35 + 39 + 41$
0%
$21 + 23 + 35 + 39 + 39 + 41$
0%
$31 + 33 + 35 + 37 + 39 + 41$
0%
$31 + 33 + 35 + 37 + 39 + 43$

Explanation

We know,

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

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

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

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

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

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

Therefore, the sum of odd consecutive cube number for

$6^3=216$ must be

$31+33+35+37+39+41$ .

Hence, option

$C$ is correct.

Find the cube root of

$0.000000027$ .

0%
$0.03$
0%
$0.3$
0%
$0.003$
0%
$0.0003$

Explanation

We know,

$0.000000027=\dfrac{27}{1000000000}$ .

Prime factorising, we get,

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

$1000000000$

$= (2\times2 \times 2) \times(2\times2\times2)\times(2\times2\times2)\times (5\times5\times5)\times(5\times5\times5)\times(5\times5\times5)$

$=2^3\times2^3\times2^3\times5^3\times5^3\times5^3$

$=1000^3$

Then, cube root of

$0.000000027=\dfrac{27}{1000000000}$ is:

$\sqrt[3]{0.000000027}=\sqrt[3]{\dfrac{27}{1000000000}}$

$=\sqrt[3]{\dfrac{3^3}{1000^3}}={\dfrac{3}{1000}}=0.003$ .

Hence, option

$C$ is correct.

How many consecutive odd numbers will be needed to obtain the sum of

$6^3$ ?

0%
$6$
0%
$5$
0%
$4$
0%
$3$

Explanation

We know,

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

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

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

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

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

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

$\displaystyle \therefore$

$6$ consecutive odd numbers are required to obtain the sum as

$6^3$ .

Hence, option

$A$ is correct.

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

$21 + 23 + 29 + 27 + 19 + 25$

0%
$21$
0%
$19$
0%
$23$
0%
$25$

Explanation

We know,

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

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

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

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

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

Therefore,

$19$ must be taken out to form a cube number,

$5^3=125$ .

Hence, option

$B$ is correct.

Find the cube root of

$238328$ by estimation method.

0%
$52$
0%
$62$
0%
$72$
0%
$182$

Explanation

Step $I$ : Form groups of $3$ starting from right most digit of $238328$ , i.e. $\displaystyle \overline {238 }\: \overline { 328 }$ .

Then, the two groups are $238$ and $328.$

Here, $238$ has $3$ digit and $328$ has $3$ digit.

Step $II$ : Take $328$ .

Digit in unit place $= 8$ .

Therefore, we take one's place of required cube root as $2$ ....[Since, $2^3=8$ ].

Step $III$ : Now, take the other group $238$ .

We know, $\displaystyle {6}^{ 3 }=216$ and ${7 }^{ 3 }=343$ .

Here, the smallest number among $6$ and $7$ is $6$ .

Therefore, we take $6$ as ten's place.

$\displaystyle \therefore \quad \sqrt[3] { 238328 } =62$ .

Hence, option $B$ is correct.

How many consecutive odd numbers will be needed to obtain the sum of

$4^3$ ?

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

Explanation

We know,

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

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

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

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

$\displaystyle \therefore$

$4$ consecutive odd numbers are required to obtain the sum as

$4^3$ .

Hence, option

$C$ is correct.

$\sqrt [3]{81x^{7}y^{10}} =$

0%
$9x^{3}y^{5}\sqrt [3]{x}$
0%
$9x^{2}y^{3}\sqrt [3]{xy}$
0%
$3x^{2}y^{3}\sqrt [3]{9xy}$
0%
$3x^{2}y^{3}\sqrt [3]{3xy}$
0%
$3x^{3}y^{5}\sqrt [3]{x}$

Which of the following statements is true?

0%
Cube root of a positive number may be a negative number.
0%
Cube root of a number ending with 8 ends with 2.
0%
Cube root of an odd number may be an even number.
0%
All above statements are false

Explanation

We know, the cube root of any positive number will always be a positive number.

Also, cube root of a number ending with

$8$ ends with

$2$ , since

$2^3=8$

Now, cube root of an odd number cannot be an even number,

Therefore, option

$B$ is correct.

By what least number must

$3600$ be divided to make it a perfect cube?

0%
$9$
0%
$50$
0%
$300$
0%
$450$

Explanation

Prime factorising $3600$ , we get,
$3600 = 36\times 100$

$3600 = 6^2 \times 10^2$

$3600 = (2\times 3)^2 \times (2 \times 5)^2$

$3600 = 3 \times 3 \times 2 \times 2 \times 2 \times 2 \times 5 \times 5$

$3600= 2^4 \times 3 ^2 \times 5^2$ .

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

Here, number of $2$ 's is $4$ , number of $3$ 's is $2$ and number of $5$ 's is $2$ .

So we need to divide $2$ , $3^2$ and $5^2$ from the factorization to make $3600$ a perfect cube.

Hence, the smallest number by which $3600$ must be divided to obtain a perfect cube is $2 \times 3^2 \times 5^2=450$ .

Therefore, option

$D$ is correct.

The cube of an

$\text{odd}$ natural number is always __________.

0%
$\text{Even}$
0%
$\text{Odd}$
0%
$\text{Even or odd}$
0%
$\text{Can't say}$

Explanation

The cube of a natural odd number is always odd.

For natural numbers, we know,

$\text{odd}\times \text{odd}=\text{odd}$

$\text{odd}\times \text{even}=\text{even}$

$\text{even}\times \text{even}=\text{even}$ .

So, cube of a natural odd number is

$\text{odd}\times \text{odd}\times \text{odd}$

$=\text{odd}\times \text{odd}$

$=\text{odd}$ .

Eg.: Cube of an odd number

$3$ is

$27$ , which is also odd.

Hence, option

$B$ is correct.

The cube root of

$27^2$ is __________.

0%
$27$
0%
$9$
0%
$3$
0%
None of these

Explanation

On prime factorising, we get,

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

Then, cube root of ${27^2}$ is:

$\sqrt[3]{27^2}=\sqrt[3]{(3^3)^2}= 3^2=9$ .

Therefore, option $B$ is correct.

The digit at units place of the cube root of

$2197$ is

0%
$3$
0%
$9$
0%
$7$
0%
$6$

Explanation

Let's check the cube of the digits

$2^{3}=8, 3^{3}=27, 4^{3}=64, 5^{3}=125, 6^{3}=216, 7^{3}=343, 8^{3}=512, 9^{3}=729$

in all these cases only 3 has a cube with unit place 7

Therefore, unit place of cube root of 2197 will be 3

Find the smallest number by which

$2340$ must be multiplied so that the product is a perfect cube.

0%
$25350$
0%
$23350$
0%
$6580$
0%
$11780$
0%
None of these

Explanation

Given number is $2340$ .

To find out,

The smallest number by which the given number should be multiplied so that the product is a perfect cube.

The given number can be represented as the product of its prime factors as:

$2340 = 2 \times 2 \times 3 \times 3 \times 5 \times 13$

$= 2 ^2 \times 3 ^2 \times 5^1 \times 13^1$ .

We know that, a perfect cube will have $3$ powers of every prime factor.

Here, number of $2$ 's is $2$ , number of $3$ 's is $2$ , number of $5$ 's is $1$ and number of $13$ 's is $1$ .

So, we need to multiply another $2,\ 3,\ 5^2$ and $13^2$ to the factorization to make $2340$ a perfect cube.

Hence, the smallest number by which $2340$ must be multiplied to obtain a perfect cube is:

$2\times 3 \times 5^2 \times 13^2$

$=25350$

Hence, option A is correct.

Ones digit of the cube of

$54$ is:

0%
$6$
0%
$4$
0%
$2$
0%
$9$

Explanation

The given number is $54$ .

Here, the ones digit is $4$ .

We know, the cube of $4$ , i.e. $4^3=64$ , whose ones digit is $4$ .

Therefore, the ones digit of the cube of $54$ is $4$ .

Hence, option $B$ is correct.

If the digits in ones place of number is

$2$ , then the ending digits of its cube will be:

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

Explanation

Given, the units digit of the number is $2$ .

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

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

Hence, option $D$ is correct.

Which of the following are cubes of even numbers?

0%
$216$
0%
$729$
0%
$512$
0%
$3375$
0%
$1000$

Explanation

We know by the rule of even numbers that the cubes of even numbers are always even.

So we look for the even numbers in the given set of numbers.

$216$ ,

$512$ ,

$1000$ are the even numbers.

Now we prime factorise

$216$ ,

$512$ ,

$1000$ :

$216 = 2\times2\times2\times3\times3\times3=2^{3}\times3^{3}=6^{3}$

$512=2\times2\times2\times2\times2\times2\times2\times2\times2=2^{3}\times2^{3}\times2^{3}=8^{3}$

$1000=2\times2\times2\times5\times5\times5\times=2^{3}\times5^{3}=10^{3}$ .

$\therefore$

$216$ ,

$512$ ,

$1000$ are cubes of even numbers.

Hence, options

$A$ ,

$C$ and

$E$ are correct.

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

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

Explanation

Prime factorisation of

$512$ :

$512=2\times2\times 2\times2\times 2\times2\times 2\times2\times 2=2^3\times 2^3\times 2^3$ .

Then,

$\sqrt[3]{512}=\sqrt[3]{(8\times8\times8)}=8$ .

$\therefore$

$8$ is the required answer.

Hence, option

$C$ is correct.

$\sqrt[3]{(125\times64)}=?$

0%
$100$
0%
$40$
0%
$20$
0%
$30$

Explanation

Prime factorisation of

$125$

$125=5\times5\times5$ .

And prime factorisation of

$64$

$64 = 2\times2\times2\times2\times2\times2=4\times4\times4$ .

$\sqrt[3]{(125\times64)}=\sqrt[3]{(5\times5\times5\times4\times4\times4)}=\sqrt[3]{(5\times5\times5)}\times\sqrt[3]{(4\times4\times4)}=5\times4=20$ .

$\therefore$

$20$ is the required answer.

Hence, option

$C$ is correct.

Write the correct answer from the given four options:
If

$m$ is the cube root of

$n$ , then

$n$ is ____.

0%
$m^3$
0%
$\sqrt{m}$
0%
$\frac{m}{3}$
0%
$\sqrt[3]{m}$

Explanation

$\sqrt[3]{n}=m$

$n=m^3$

Given,

$m$ is the cube root of

$n$ , i.e.

$\sqrt[3]{n}=m$ .

Then,

$n=m\times m\times m=m^3$ .

Thus,

$n$ is

$m^3$ .

Hence, option

$A$ is correct.

Write the correct answer from the given four options.
Which of the following numbers is not a perfect cube?

0%
$216$
0%
$567$
0%
$125$
0%
$343$

Explanation

On prime factorising, we get,

$567 = \underline{3 \times 3 \times 3} \times 3 \times 7$ .

Here,

$3$ and

$7$ does not occur as triplets.

Hence,

$567$ is not a perfect cube.

Also,

$216 = \underline{6 \times 6\times 6}$ ,

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

and

$343 = \underline{7 \times 7\times 7}$ .

Here, each prime factors occur as triplets and therefore, they are perfect cubes.

Hence, option

$B$ is correct.

1649203_1791787_ans_da25776b3b0b462db9923a681443d3a4.png

Choose the correct answer from the given four options:
Which of the following numbers is a perfect cube?

0%
$243$
0%
$216$
0%
$392$
0%
$8640$

Explanation

Prime factorising of:

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

B) $216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3$ $= 2^3 \times 3^3$ .

C) $392 = 2 \times 2 \times 2 \times 7 \times 7$ $= 2^3 \times 7^2$ .

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

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

Here, only $216$ satisfies this condition.

Therefore, option $B$ is correct.

Write the correct answer from the given four options:
The one's digit of the cube of

$23$ is ______.

0%
$6$
0%
$7$
0%
$3$
0%
$9$

Explanation

Given, the number is $23$ .

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 $23$ is $7$ .

Hence, option $B$ is correct.

State whether the given statement is true (T) or false (F).
The cube root of

$8000$ is

$200$ .

0%
True
0%
False

Explanation

Prime factorisation of

$8000$ :

${8000} =(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (5 \times 5 \times 5) = 2^3 \times 2^3 \times 5^3$ .

Therefore,

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

Hence,

$20$ is the required answer.

That is, the given statement is false.

Thus, option

$B$ is correct.

State whether the statements are true (T) or false (F).
There are five perfect cubes between 1 and 100.

0%
True
0%
False

State whether the statements are true (T) or false (F).
If

$a^2$ ends in

$5$ , then

$a^3$ ends in

$25$ .

0%
True
0%
False

Explanation

Let us consider an example, i.e. let

$a=55$ .

$\because (55)^2 = 3025$ ends in 5 and

$(55)^3 = 166375$ does not ends in 25.

Therefore, the given statement is not true, i.e. false.

Hence, option

$B$ is correct.

State whether the statements are true (T) or false (F).
The cube of a one digit number cannot be a two digit number.

0%
True
0%
False

Explanation

No, the cube of a one digit number can be a two digit number.

For example, consider the one-digit numbers

$1,2,3,4$ .

Then, their cubes are

$1^3=1, 2^3=8, 3^3=27, 4^3=64$ .

Clearly, the cube of a one digit number can be a two digit number.

Hence, the given statement is false.

State whether the statements are true (T) or false (F).
Cube of an even number is odd.

0%
True
0%
False

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 an even natural numbers is even.

Therefore, the given statement is false and option

$B$ is correct.

CBSE Questions for Class 8 Maths Cubes And Cube Roots Quiz 7 - 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 7 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

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