Cubes and Cube Roots Test

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Cubes and Cube Roots Test - 12

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Weekly Quiz Competition

Question 1
1 / -0

The smallest natural number by which $$36$$ must be multiplied to get a perfect cube is _____.

A
$$6$$

B
$$216$$

C
$$45$$

D
$$2$$

Solution

Prime factorising $$36$$, we get,
$$36=2 \times 2 \times 3 \times 3=2^2 \times 3^2$$.
We know, a perfect cube has multiples of $$3$$ as powers of prime factors.
Here, number of $$2$$'s is $$2$$ and number of $$3$$'s is $$2$$.
So we need to multiply another $$2$$ and $$3$$ in the factorization to make $$36$$ a perfect cube.
Hence, the smallest number by which $$36$$ must be multiplied to obtain a perfect cube is $$2\times 3 =6$$.
Hence, option $$A$$ is correct.
Question 2
1 / -0

Choose the correct option:
There is no perfect cube which ends in $$4$$.

A
True

B
False

C
Ambiguous

D
Data insufficient

Solution

We know,
cube of $$4$$, i.e. $$4^3=64$$, which is a perfect cube.
That is, there exists a perfect cube which ends in $$4$$.
Therefore, the given statement is false and option $$B$$ is correct.
Question 3
1 / -0

Value of $$\sqrt [3]{0.008}$$ is:

A
$$0.2$$

B
$$0.4$$

C
$$0.002$$

D
None

Solution

We know,
$$0.008 = 8 \times { 10 }^{ -3 }=8 \times \dfrac {1}{10} \times \dfrac{1}{10} \times \dfrac {1}{10}$$.

Also, on prime factorising $$8$$, we get,
$$8=2\times 2\times 2=2^3$$.

Then, $$\sqrt [ 3 ]{ 0.008 } $$ $$=\sqrt[3]{8 \times \dfrac {1}{10} \times \dfrac{1}{10} \times \dfrac {1}{10}}$$
$$=\sqrt[3]{2^3 \times \dfrac {1}{10} \times \dfrac{1}{10} \times \dfrac {1}{10}}$$
$$=2\times \dfrac {1}{10}=0.2$$.

Hence, option $$A$$ is correct.
Question 4
1 / -0

The cube of $$-3.1$$ is:

A
$$ -29.791$$

B
$$-2.6891$$

C
$$-2.5781$$

D
None

Solution

Cube of the number $$-3.1$$:
$$(-3.1)^{ 3 } = -3.1\times-3.1\times -3.1 = -29.791$$.

Hence, option $$A$$ is correct.
Question 5
1 / -0

Find the cube of $$0.6$$.

A
$$0.216$$

B
$$0.36$$

C
$$21.6$$

D
$$2.16$$

Solution

Cube of $$0.6$$ is:
$$(0.6)^3=0.6\times 0.6\times 0.6$$
$$=0.216$$.

Hence, option $$A$$ is correct.
Question 6
1 / -0

Find the smallest number by which the following number must be multiplied to obtain a perfect cube:
$$243$$

A
$$3$$

B
$$1$$

C
$$0$$

D
$$5$$

Solution

Prime factorizing $$243$$, we get,
$$243=3\times 3\times 3\times 3\times 3=3^5$$.

We know, a perfect cube has multiples of $$3$$ as powers of prime factors.
Here, number of $$3$$'s is $$5$$.
So we need to multiply another $$3$$ in the factorization to make $$243$$ a perfect cube.

Hence, the smallest number by which $$243$$ must be multiplied to obtain a perfect cube is $$3$$.

Therefore, option $$A$$ is correct.
Question 7
1 / -0

Find $$\sqrt[3]{125}.$$

A
$$10$$

B
$$5$$

C
$$-7$$

D
$$15$$

Solution

On prime factorising, we get,
$$125=5 \times 5\times 5$$ $$= 5^3$$.
Then, cube root of $$125$$ is:
$$\sqrt[3]{125}=\sqrt[3]{5^3}= 5$$.
Therefore, option $$B$$ is correct.
Question 8
1 / -0

Find the smallest number by which the following number must be divided to obtain a perfect cube.
192

A
3

B
4

C
7

D
1

Solution

$$Factorization\quad of\quad 192$$

$$192\quad =\quad 2\times2\times2\times2\times2\times2\times3$$

$$=\quad 2^{ 6 }\times3$$

$$To\quad make\quad a\quad perfect\quad cube,\quad we\quad need\quad to\quad have\quad multiples\quad of\quad 3\quad a\quad powers\quad of\quad prime\quad factors,\\ i.e,\quad we\quad divide\quad the\quad number\quad by\quad 3$$
Question 9
1 / -0

Value of $$ \displaystyle \sqrt[3]{343}$$ is:

A
$$7$$

B
$$-5$$

C
$$ \displaystyle \frac{7}{5}$$

D
$$ \displaystyle \frac{5}{7}$$

Solution

On prime factorising, we get,
$${343}$$$$= {7\times 7\times 7}$$ $$=7^3$$.

Then,
$$\sqrt [3] {343}$$$$=\sqrt [3] {7^3}$$ $$=7$$.

Hence, option $$A$$ is correct.
Question 10
1 / -0

Find the smallest number by which the following number must be divided to obtain a perfect cube.
704

A
11

B
12

C
14

D
15

Solution

$$Factorization\quad of\quad 704$$

$$ 704\quad =\quad 2\times2\times2\times2\times2\times2\times11$$

$$=\quad 2^{ 6 }\times11$$

$$To\quad make\quad a\quad perfect\quad cube,\quad we\quad need\quad to\quad have\quad multiples\quad of\quad 3\quad a\quad powers\quad of\quad prime\quad factors,\\ i.e,\quad we\quad divide\quad the\quad number\quad by\quad 11$$

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Re-Attempt Weekly Quiz Competition

Cubes and Cube Roots Test - 12

Result

Cubes and Cube Roots Test - 12

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SHARING IS CARING

Question 1

$$6$$

$$216$$

$$45$$

$$2$$

Solution

Question 2

True

False

Ambiguous

Data insufficient

Solution

Question 3

$$0.2$$

$$0.4$$

$$0.002$$

None

Solution

Question 4

$$ -29.791$$

$$-2.6891$$

$$-2.5781$$

None

Solution

Question 5

$$0.216$$

$$0.36$$

$$21.6$$

$$2.16$$

Solution

Question 6

$$3$$

$$1$$

$$0$$

$$5$$

Solution

Question 7

$$10$$

$$5$$

$$-7$$

$$15$$

Solution

Question 8

3

4

7

1

Solution

Question 9

$$7$$

$$-5$$

$$ \displaystyle \frac{7}{5}$$

$$ \displaystyle \frac{5}{7}$$

Solution

Question 10

11

12

14

15

Solution

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