MCQ Questions for Class 8 Maths Cubes And Cube Roots Quiz 5

Find the smallest numbers by which

$11979$ must be multiplied so that the product is a perfect cube.

0%
$1$
0%
$1.01$
0%
$3.01$
0%
$3$
0%
$4.01$
0%
$5.01$
0%
$4$

Explanation

Prime factorising $11979$ , we get,

$11979 = 3 \times 3 \times 11 \times 11 \times 11$

$= 3^2 \times 11 ^3$ .

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

Here, number of $3$ 's is $2$ and number of $11$ 's is $3$ .

So we need to multiply another $3$ to the factorization to make $11979$ a perfect cube.

Hence, the smallest number by which $11979$ must be multiplied to obtain a perfect cube is $3$ .

Hence, option $D$ is correct.

Write the correct answer from the given four options.

$\sqrt[3]{1000}$ is equal to ______.

0%
$10$
0%
$100$
0%
$1$
0%
None of these

Explanation

Prime factorisation of

$1000$ :

$1000=2 \times 2 \times 2 \times 5 \times 5 \times 5=10 \times 10 \times 10$ .

Therefore,

$\sqrt[3]{1000} = \sqrt{10 \times 10 \times 10} = 10$ .

Hence,

$10$ is the required answer.

Thus, option

$A$ is correct.

What is the value of

$\displaystyle \sqrt[3]{\left (64\times 729 \right )}=$ ?

0%
$9$
0%
$6$
0%
$36$
0%
$52$

Explanation

On prime factorisation of the numbers individually, we get,

$64=\underline { 2 \times 2 \times 2 } \times \underline { 2 \times 2 \times 2 }$ $=2^3 \times 2^3=4^3$ .

$729=\underline { 3 \times 3 \times 3 } \times \underline { 3 \times 3\times 3 }$ $=3^3 \times 3^3=9^3$ .

Therefore,

$\sqrt [ 3 ]{ 64 \times 729 }$ $= \sqrt [ 3 ]{4^3 \times 9^3 }$ $=4\times9=36$ .

Hence, option $C$ is correct.

$\displaystyle \sqrt[3]{(12167 \times 1728)} =$ ?

0%
$276$
0%
$318$
0%
$216$
0%
$356$

Explanation

Given, to find $\sqrt[3]{12167\times 1728}$ .

Prime factorising, we get,

$12167= 23\times23\times23=23^3$

and $1728$ $= 2\times2\times2\times2\times2\times2\times3\times3\times3 \\ =2^3 \times 2^3 \times 3^3.$

Then,

$\sqrt[3]{12167\times 1728}$

$= \sqrt[3]{23^3}\times \sqrt[3]{2^3 \times 2^3 \times 3^3}$ .

$= 23\times2\times2\times 3$

$= 276$ .

Hence, option $A$ is correct.

Simplify:

$\sqrt[3]{\dfrac {216}{2197}}$ .

0%
$\dfrac{36}{27}$
0%
$\dfrac{66}{23}$
0%
$\dfrac{6}{13}$
0%
None of the above

Explanation

On prime factorisation of the numbers individually, we get,

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

$2197=\underline { 13 \times 13 \times 13 } =13^3$ .

Therefore,

$\ \sqrt[3] {\cfrac {216}{2197}}$ $= \cfrac {\sqrt[3]{216}} {\sqrt[3]{2197}}$ $=\cfrac {\sqrt[3]{ 6^3}}{ \sqrt[3]{13^3}}$ $=\dfrac {6}{13}.$

Thus, option $C$ is correct.

Find the smallest number by which

$243$ must be multiplied, so that the product is a perfect cube.

0%
$6$
0%
$3$
0%
$7$
0%
none of these

Explanation

$\displaystyle 243=3\times 3\times 3\times 3\times 3$

$\displaystyle =\left( { 3 }^{ 3 }\times { 3 }^{ 2 } \right) ={ \left( 3 \right) }^{ 3 }\times { 3 }^{ 2 }$
In this factorization there is no triplet for

$3.$
So,

$243$ is not a perfect cube.

$\displaystyle \therefore$

$243$ has to be multiplied by

$3$ to make it a perfect cube.

Write the units digit of the cube for

$388$ .

0%
$4$
0%
$3$
0%
$2$
0%
$8$

Explanation

Given, the number is $388$ .

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 $388$ is $2$ .

Hence, option $C$ is correct.

Write the units digit of the cube for

$109$ .

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

Explanation

Given, the number is $109$ .

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 $109$ is $9$ .

Hence, option $C$ is correct.

Which of the following numbers are not perfect cubes?

0%
$64$
0%
$729$
0%
$243$
0%
$81$

Explanation

We know, a perfect cube has multiples of

$3$ as powers of prime factors.

On prime factorisation of the numbers individually, we get,

A) $64=\underline { 2 \times 2 \times 2 } \times \underline { 2 \times 2 \times 2 }=4^3$ .

B) $729=\underline { 3 \times 3 \times 3 } \times \underline { 3\times 3 \times 3 }=3^3 \times 3^3=9^3$ .

C) $243=\underline { 3 \times 3 \times 3 } \times { 3 \times 3 }=3^5$ .

D) $81=\underline { 3 \times 3 \times 3 } \times { 3 }=3^4$ .

Here, clearly, options

$C$ and

$D$ are not perfect cubes.

Write the units digit of the cube for

$77774$ .

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

Explanation

Given, the number is $77774$ .

Here, the units digit is $4$ .

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

Therefore, the units digit of the cube of $77774$ is $4$ .

Hence, option $A$ is correct.

Find the cube of: 819

0%
$549353259$
0%
$549353258$
0%
$549353249$
0%
$549353250$

Explanation

$\displaystyle 819=3\times 3\times 7\times 13$

$\Rightarrow$

$\displaystyle{ \left( 819 \right) }^{ 3 }={ \left( 3 \right) }^{ 3 }\times { \left( 3 \right) }^{ 3 }\times { \left( 7 \right) }^{ 3 }\times { \left( 13 \right) }^{ 3 }$

$\displaystyle =3\times 3\times 3\times 3\times 3\times 3\times 7\times 7\times 7\times 13\times 13\times 13$

$\displaystyle =27\times 27\times 343\times 2197$

$\displaystyle =549353259$

$\displaystyle \therefore$ The cube of

$819 = 549353259.$

Find the smallest number by which

$128$ must be divided, so that the quotient is a perfect cube.

0%
$2$
0%
$3$
0%
$7$
0%
$12$

Explanation

$\displaystyle 128=2\times 2\times 2\times 2\times 2\times 2\times 2$

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

$\displaystyle ={ \left( 2\times 2 \right) }^{ 3 }\times 2$
In the above factorization there is no triplet for

$2$ .
So,

$128$ is not a perfect cube.
Therefore,

$\displaystyle$

$128$ must be divided by

$2$ to make the quotient a perfect cube.

Find the smallest number by which

$72$ must be multiplied, so that the product is a perfect cube.

0%
$3$
0%
$6$
0%
$12$
0%
$4$

Explanation

$\displaystyle 72=2\times 2\times 2\times 3\times 3$

$\displaystyle =\left( { 2 }^{ 3 }\times { 3 }^{ 2 } \right) ={ \left( 2 \right) }^{ 3 }\times 3^{2}$
In this factorisation there is no triplet for

$3$ .
So,

$72$ is not a perfect cube.
Therefore,

$\displaystyle$

$72$ has to be multiplied by

$3$ to make it a perfect cube.

Find the cube root of the following number:

$274625000$

0%
$650$
0%
$625$
0%
$750$
0%
$450$

Explanation

On prime factorising, we get,

$274625000=(5 \times 5\times 5)\times (5\times5\times5)\times(2\times2\times2)\times(13\times13\times13)$

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

Then, cube root of $274625000$ is:

$\sqrt[3]{274625000}=\sqrt[3]{5^3 \times 5^3 \times 2^3 \times 13^3}= 5\times 5\times 2\times 13=650$ .

Therefore, option $A$ is correct.

$130$

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

Explanation

$130-1=129, 129-7=122, 122-19=103, 103-37=66, 66-61=5.$

$\displaystyle \Rightarrow$ The remainder got is

$5.$

$\displaystyle \Rightarrow$

$5$ is the number to be subtracted from

$130$ to make it a perfect cube.

$\displaystyle \Rightarrow$

$130-5=125$ is the perfect cube.

$\displaystyle \therefore$ The corresponding cube

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

Find the cube root of

$389017$ by finding their units and ten digits.

0%
$63$
0%
$67$
0%
$73$
0%
$77$

Explanation

Given,

$389017$ .

Here, the unit digit is

$7.$

$\therefore$ The unit digit of its cube root is

$3$ .....

$[\because 3^3=27]$ .

After grouping the last three digits from the right, the number left is

$389.$

Now,

$7^3=343<389$ and

$8^3=512>389.$

$\therefore$ The tens digit of the cube root is

$7.$

Therefore,

$\sqrt[3]{389017}=73$ .

Hence, option

$C$ is correct.

Find the smallest number by which

$192$ must be divided, so that the quotient is a perfect cube.

0%
$2$
0%
$3$
0%
$4$
0%
$7$

Explanation

By prime factorisation method, we have

$\displaystyle 192=2\times 2\times 2\times 2\times 2\times 2\times 3$

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

$\displaystyle ={ \left( 2\times 2 \right) }^{ 3 }\times 3$
In the above factorization there is no triplet for

$3$ .
So,

$192$ is not a perfect cube.
Therefore,

$192$ must be divided by

$3$ to make the quotient a perfect cube.

Find the smallest number by which

$135$ must be divided, so that the quotient is a perfect cube.

0%
$3$
0%
$5$
0%
$9$
0%
$15$

Explanation

$\displaystyle 135=5\times 3\times 3\times 3$

$\displaystyle =\left( 5\times { 3 }^{ 3 } \right)$

$\displaystyle ={ \left( 3 \right) }^{ 3 }\times 5$
In the above factorization there is triplet for

$5$ .
So,

$135$ is not a perfect cube.

$\displaystyle \therefore$

$135$ must be divided by

$5$ to make the quotient a perfect cube.

Find the cube root of the following number by finding their units and ten digits:

$46656$

0%
$36$
0%
$31$
0%
$32$
0%
$35$

Explanation

Here the unit digit is $$6.$$

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

$6.$

$\because 6^3=36$

After striking out the last three digits from the right, the number left is

$46.$

Now,

$3^3=27<46$ and

$4^3=64>46$

The ten-digit of the cube root is

$3.$

$\therefore \sqrt[3]{46656}=36$

$\displaystyle { a }^{ 2 }$ ends in an even number of zeros, then

$\displaystyle { a }^{ 3 }$ ends in an odd number of zeros.

0%
True
0%
False
0%
Ambiguous
0%
Insufficient information

Explanation

False,

$100 \times 100$

$= 10000$ ends in even number of zeros.

$100 \times 100 \times 100 = 1000000$ also ends in even number of zeros.

So the statement if

$\displaystyle { a }^{ 2 }$ ends in an even number of zeros, then

$\displaystyle { a }^{ 3 }$ ends in an odd number of zeros is not true for all cases.

Find the cube root

$64$ by successive subtraction of numbers:

$(1, 7 ,19, 37, 61, 91, 127, 169, 217, 271, 331, 397, ...)$

0%
$6$
0%
$12$
0%
$8$
0%
$4$

Explanation

By consecutive subtraction of numbers

$1,7,19,37,61,91,127,...$ from

$64$ , we get,

$64-1=63, 63-7=56, 56-19=37, 37-37=0.$
That is, we obtain the difference is

$0$ after

$4$ successive subtractions.

$\displaystyle \therefore$ The cube root of

$64=4$ .

Hence, option

$D$ is correct.

$\displaystyle { a }^{ 2 }$ ends in

$9,$ then

$\displaystyle { a }^{ 3 }$ will end in:

0%
$7$
0%
$3$
0%
$5$
0%
none of the above

Explanation

Given that,

$a^2$ ends in

$9$ .

To find out: What will

$a^3$ end in?

$a^2$ ends in

$9$

$\therefore \ a$ will either end in

$3$ or

$7$ .

Case I: If

$a$ ends in

$3$ :

$\therefore \ a^3$ will end in

$7\quad \quad [\because \ 3\times 3\times 3=2\underline7]$ .

Case II: If

$a$ ends in

$7$ :

$\therefore \ a^3$ will end in

$3\quad \quad [\because \ 7\times 7\times 7=34\underline3]$ .

Hence, both A and B are correct.

Find the cube root of

$614125$ using prime factorization:

0%
$65$
0%
$75$
0%
$85$
0%
$95$

Explanation

On prime factorising, we get,

$614125 = 5\times5\times5\times17\times17\times17$

$= 5^3\times17^3$ .

Then, cube root of $614125$ is:

$\sqrt[3]{614125}=\sqrt[3]{5^3 \times 17^3}= 5\times17=85$ .

Therefore, option $C$ is correct.

$-571787$

0%
$-73$
0%
$-83$
0%
$-93$
0%
$-63$

Explanation

On prime factorising, we get,

$571787=\underline{83\times 83\times 83}$ $=83^3$ .

Then, $-571787$ $=(-83)^3$ .

Therefore, cube root of ${-571787}$ is:

$\sqrt[3]{-571787}=\sqrt[3]{(-83)^3}= -83$ .

Therefore, option $B$ is correct.

Find the cube root of:

$175616$ .

0%
$56$
0%
$46$
0%
$66$
0%
$76$

Explanation

On prime factorising, we get,

$175616= 8\times8\times8\times7\times7\times7$

$= 8^3 \times 7^3$ .

Then, cube root of $175616$ is:

$\sqrt[3]{175616}=\sqrt[3]{8^3 \times 7^3}= 8\times 7\times=56$ .

Therefore, option $A$ is correct.

By estimation method,

$\sqrt[3]{-13824}=$ ?

0%
$-24$
0%
$-28$
0%
$-26$
0%
$-34$

Explanation

Given,

$-13824$ .
The unit's digit of

$13824$ is

$4,$ so

$4$ is the unit's digit of the cube root....[Since,

$4^3=64$ ].
Now, grouping the three digits from the right, the number left is

$13.$
We know,

$\displaystyle { 2 }^{ 3 }=8<13<27={ 3 }^{ 3 }$ .

$\displaystyle \Rightarrow$

$2$ is the ten's digit of the cube root of

$13824.$

Then,

$24$ is the cube root of

$13824.$

$\displaystyle \therefore$

$-24$ is the cube root of

$-13824.$

Hence, option

$A$ is correct.

Find the cube root of

$592704$ using prime factorization:

0%
$86$
0%
$82$
0%
$84$
0%
$88$

Explanation

On prime factorizing, we get,

$592704 = 2\times2\times2\times2\times2\times2\times3\times3\times3\times7\times7\times7$

$= 2^3\times2^3\times3^3\times7^3 \\=84^3.$

Thus, cube root of $592704$ is:

$\sqrt[3]{592704}=\sqrt[3]{84^3 } \\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =84$ .

Therefore, option $C$ is correct.

Find the cube root of

$250047$ using prime factorization:

0%
$63$
0%
$61$
0%
$67$
0%
$65$

Explanation

On prime factorising, we get,

$250047$ $=\underline{3\times3\times3}\times\underline{3\times3\times3}\times\underline{7\times 7\times 7}$

$= 3^3\times3^3\times7^3=63^3$ .

Then, cube root of $250047$ is:

$\sqrt[3]{250047}=\sqrt[3]{63^3}= 63$ .

Therefore, option $A$ is correct.

$\sqrt[3]{-226981}=$ ?

0%
$-51$
0%
$-61$
0%
$-67$
0%
$-57$

Explanation

On prime factorising, we get,

$226981=\underline{61\times 61\times 61}$ $=61^3$ .

Then, $-226981$ $=(-61)^3$ .

Therefore, value of $\sqrt[3]{-226981}$ is:

$\sqrt[3]{-226981}=\sqrt[3]{(-61)^3}= -61$ .

Therefore, option $B$ is correct.

Find the cube root of

$438976$ using prime factorization:

0%
$74$
0%
$72$
0%
$76$
0%
$71$

Explanation

On prime factorising, we get,

$438976$ $=\underline{2\times2\times2}\times\underline{2\times2\times2}\times\underline{19\times 19\times 19}$

$= 2^3\times2^3\times19^3=76^3$ .

Then, cube root of $438976$ is:

$\sqrt[3]{438976}=\sqrt[3]{76^3}= 76$ .

Therefore, option $C$ is correct.

CBSE Questions for Class 8 Maths Cubes And Cube Roots Quiz 5 - 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 5 - MCQExams.com

0:0:1

Practice Class 8 Maths Quiz Questions and Answers

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