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

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.

We know,

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

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

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

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

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

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.

On prime factorising, we get,

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.

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.

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.

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.

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.