Cubes and Cube Roots Test

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Cubes and Cube Roots Test - 25

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Weekly Quiz Competition

Question 1
1 / -0

Which number should be multiplied by $$72$$ to make it a perfect cube?

A
$$3$$

B
$$2$$

C
$$9$$

D
$$8$$

Solution

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 $$2^3X3^3$$.
Question 2
1 / -0

Find the smallest number by which $$500$$ should be multiplied to make it a perfect cube.

A
$$5$$

B
$$10$$

C
$$3$$

D
$$2$$

Solution

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.
Question 3
1 / -0

Cube root of $$0.000216$$ is:

A
$$0.6$$

B
$$0.06$$

C
$$0.006$$

D
None of the above

Solution

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.
Question 4
1 / -0

How many consecutive odd numbers are needed to obtain sum as $$\displaystyle { 3 }^{ 3 }$$?

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

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.
Question 5
1 / -0

Find the cube root of $$166375$$ by prime factorization method.

A
$$45$$

B
$$55$$

C
$$65$$

D
$$75$$

Solution

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.
Question 6
1 / -0

Find the cube root of $$39304$$ by estimation method.

A
$$24$$

B
$$44$$

C
$$34$$

D
$$54$$

Solution

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.
Question 7
1 / -0

The cube root of any multiple of $$8$$ is always divisible by:

A
$$2$$

B
$$4$$

C
$$8$$

D
$$16$$

Solution

Prime factorising $$8$$, we get,
$$8 = 2\times 2\times 2=2^3$$.

If $$x$$ 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.
Question 8
1 / -0

Cube of $$1.5$$ is _________.

A
$$3.375$$

B
$$33.75$$

C
$$3.125$$

D
$$31.25$$

Solution

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.
Question 9
1 / -0

If the unit place of a number is $$7$$, then find the unit digit of its cube.

A
$$1$$

B
$$3$$

C
$$5$$

D
$$7$$

Solution

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.
Question 10
1 / -0

Find the cube root of $$59319$$ by estimation method.

A
$$29$$

B
$$39$$

C
$$49$$

D
$$59$$

Solution

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.

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Cubes and Cube Roots Test - 25

Result

Cubes and Cube Roots Test - 25

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Question 1

$$3$$

$$2$$

$$9$$

$$8$$

Solution

Question 2

$$5$$

$$10$$

$$3$$

$$2$$

Solution

Question 3

$$0.6$$

$$0.06$$

$$0.006$$

None of the above

Solution

Question 4

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 5

$$45$$

$$55$$

$$65$$

$$75$$

Solution

Question 6

$$24$$

$$44$$

$$34$$

$$54$$

Solution

Question 7

$$2$$

$$4$$

$$8$$

$$16$$

Solution

Question 8

$$3.375$$

$$33.75$$

$$3.125$$

$$31.25$$

Solution

Question 9

$$1$$

$$3$$

$$5$$

$$7$$

Solution

Question 10

$$29$$

$$39$$

$$49$$

$$59$$

Solution

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