Cubes and Cube Roots Test

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Cubes and Cube Roots Test - 29

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Weekly Quiz Competition

Question 1
1 / -0

The cube of an $$\text{odd}$$ natural number is always __________.

A
$$\text{Even}$$

B
$$\text{Odd}$$

C
$$\text{Even or odd}$$

D
$$\text{Can't say}$$

Solution

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.
Question 2
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The cube root of $$27^2$$ is __________.

A
$$27$$

B
$$9$$

C
$$3$$

D
None of these

Solution

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.
Question 3
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The digit at units place of the cube root of $$2197$$is

A
$$3$$

B
$$9$$

C
$$7$$

D
$$6$$

Solution

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
Question 4
1 / -0

Find the smallest number by which $$2340$$ must be multiplied so that the product is a perfect cube.

A
$$25350$$

B
$$23350$$

C
$$6580$$

D
$$11780$$

E
None of these

Solution

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.
Question 5
1 / -0

Ones digit of the cube of $$54$$ is:

A
$$6$$

B
$$4$$

C
$$2$$

D
$$9$$

Solution

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.
Question 6
1 / -0

If the digits in ones place of number is $$2$$, then the ending digits of its cube will be:

A
$$2$$

B
$$4$$

C
$$6$$

D
$$8$$

Solution

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.
Question 7
1 / -0

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

A
$$6$$

B
$$7$$

C
$$8$$

D
$$9$$

Solution

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.
Question 8
1 / -0

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

A
$$243$$

B
$$216$$

C
$$392$$

D
$$8640$$

Solution

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.
Question 9
1 / -0

Write the correct answer from the given four options:
The one's digit of the cube of $$23$$ is ______.

A
$$6$$

B
$$7$$

C
$$3$$

D
$$9$$

Solution

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.
Question 10
1 / -0

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

A
$$100$$

B
$$40$$

C
$$20$$

D
$$30$$

Solution

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.

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Re-Attempt Weekly Quiz Competition

Cubes and Cube Roots Test - 29

Result

Cubes and Cube Roots Test - 29

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SHARING IS CARING

Question 1

$$\text{Even}$$

$$\text{Odd}$$

$$\text{Even or odd}$$

$$\text{Can't say}$$

Solution

Question 2

$$27$$

$$9$$

$$3$$

None of these

Solution

Question 3

$$3$$

$$9$$

$$7$$

$$6$$

Solution

Question 4

$$25350$$

$$23350$$

$$6580$$

$$11780$$

None of these

Solution

Question 5

$$6$$

$$4$$

$$2$$

$$9$$

Solution

Question 6

$$2$$

$$4$$

$$6$$

$$8$$

Solution

Question 7

$$6$$

$$7$$

$$8$$

$$9$$

Solution

Question 8

$$243$$

$$216$$

$$392$$

$$8640$$

Solution

Question 9

$$6$$

$$7$$

$$3$$

$$9$$

Solution

Question 10

$$100$$

$$40$$

$$20$$

$$30$$

Solution

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