Cubes and Cube Roots Test

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Cubes and Cube Roots Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

Find the cube root of $$24389$$ by estimation method.

A
$$27$$

B
$$31$$

C
$$29$$

D
$$33$$

Solution

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.
Question 2
1 / -0

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

A
$$\displaystyle { 122 }^{ 3 }$$

B
$$\displaystyle { 168 }^{ 3 }$$

C
$$\displaystyle { 137 }^{ 3 }$$

D
$$\displaystyle { 184 }^{ 3 }$$

Solution

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.
Question 3
1 / -0

Find the cube root of $$0.000000027$$.

A
$$0.03$$

B
$$0.3$$

C
$$0.003$$

D
$$0.0003$$

Solution

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.
Question 4
1 / -0

How many consecutive odd numbers will be needed to obtain the sum of $$6^3$$?

A
$$6$$

B
$$5$$

C
$$4$$

D
$$3$$

Solution

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.
Question 5
1 / -0

Which odd number needs to be taken out to form a cube number?
$$21 + 23 + 29 + 27 + 19 + 25$$

A
$$21$$

B
$$19$$

C
$$23$$

D
$$25$$

Solution

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.
Question 6
1 / -0

Find the cube root of $$238328$$ by estimation method.

A
$$52$$

B
$$62$$

C
$$72$$

D
$$182$$

Solution

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.
Question 7
1 / -0

How many consecutive odd numbers will be needed to obtain the sum of $$4^3$$?

A
$$2$$

B
$$3$$

C
$$4$$

D
$$5$$

Solution

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.
Question 8
1 / -0

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

A
$$9x^{3}y^{5}\sqrt [3]{x}$$

B
$$9x^{2}y^{3}\sqrt [3]{xy}$$

C
$$3x^{2}y^{3}\sqrt [3]{9xy}$$

D
$$3x^{2}y^{3}\sqrt [3]{3xy}$$

E
$$3x^{3}y^{5}\sqrt [3]{x}$$

Solution
Question 9
1 / -0

Which of the following statements is true?

A
Cube root of a positive number may be a negative number.

B
Cube root of a number ending with 8 ends with 2.

C
Cube root of an odd number may be an even number.

D
All above statements are false

Solution

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.
Question 10
1 / -0

By what least number must $$3600$$ be divided to make it a perfect cube?

A
$$9$$

B
$$50$$

C
$$300$$

D
$$450$$

Solution

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.

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Cubes and Cube Roots Test - 28

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Cubes and Cube Roots Test - 28

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Question 1

$$27$$

$$31$$

$$29$$

$$33$$

Solution

Question 2

$$\displaystyle { 122 }^{ 3 }$$

$$\displaystyle { 168 }^{ 3 }$$

$$\displaystyle { 137 }^{ 3 }$$

$$\displaystyle { 184 }^{ 3 }$$

Solution

Question 3

$$0.03$$

$$0.3$$

$$0.003$$

$$0.0003$$

Solution

Question 4

$$6$$

$$5$$

$$4$$

$$3$$

Solution

Question 5

$$21$$

$$19$$

$$23$$

$$25$$

Solution

Question 6

$$52$$

$$62$$

$$72$$

$$182$$

Solution

Question 7

$$2$$

$$3$$

$$4$$

$$5$$

Solution

Question 8

$$9x^{3}y^{5}\sqrt [3]{x}$$

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

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

$$3x^{2}y^{3}\sqrt [3]{3xy}$$

$$3x^{3}y^{5}\sqrt [3]{x}$$

Solution

Question 9

Cube root of a positive number may be a negative number.

Cube root of a number ending with 8 ends with 2.

Cube root of an odd number may be an even number.

All above statements are false

Solution

Question 10

$$9$$

$$50$$

$$300$$

$$450$$

Solution

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