Cubes and Cube Roots — Multiple Choice Questions

Cube of a negative number is

1. negative

2. positive

3. negative or positive

4. none of these.

Answer

The cube of a negative integer is always negative.

Let the negative number be $(-a)$, where $a$ is a positive number.

Solving,

$\Rightarrow (-a)^3\\[1em] \Rightarrow (-a) \times (-a) \times (-a)\\[1em] \Rightarrow -a^3$

Since $a^3$ is positive, so -$a^3$ is negative.

Hence, Option 1 is the correct option.

The unit digit of cube of 476 is

1. 4

2. 6

3. 8

4. 2

Answer

If a number has 6 in the unit place, then its cube also ends in 6.

One's digit of 476 = 6

Cube of one's digit = 6 × 6 × 6 = 216

Hence, Option 2 is the correct option.

Cube of (-8) is

1. -512

2. 512

3. -64

4. 64

Answer

Solving,

$\Rightarrow (-8)^3\\[1em] \Rightarrow (-8) \times (-8) \times (-8)\\[1em] \Rightarrow -512$

Hence, Option 1 is the correct option.

Cube of $\left(\dfrac{-3}{7}\right)$ is

1. $\dfrac{27}{343}$

2. $\dfrac{-27}{343}$

3. $\dfrac{9}{49}$

4. $\dfrac{-9}{49}$

Answer

Solving,

$\Rightarrow \left(\dfrac{-3}{7}\right)^3\\[1em] \Rightarrow \dfrac{(-3) \times (-3) \times (-3)}{7 \times 7 \times 7}\\[1em] \Rightarrow \dfrac{-27}{343}$

Hence, Option 2 is the correct option.

Cube root of -1331 is

1. 11

2. 21

3. -11

4. -21

Answer

Finding prime factors of 1331 :

$\begin{array}{l|r} 11 & 1331 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \\ \end{array}$

Expressing 1331 into prime factors, we have

1331 = 11 × 11 × 11

$\Rightarrow \sqrt[3]{-1331}\\[1em] \Rightarrow -\sqrt[3]{1331}\\[1em] \Rightarrow -\sqrt[3]{11 \times 11 \times 11}\\[1em] \Rightarrow -11$

Hence, Option 3 is the correct option.

Cube root of 2744 is

1. 16

2. 18

3. -14

4. 14

Answer

Finding prime factors of 2744 :

$\begin{array}{l|r} 2 & 2744 \\ \hline 2 & 1372 \\ \hline 2 & 686 \\ \hline 7 & 343 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$

Expressing 2744 into prime factors, we have

2744 = (2 × 2 × 2) × (7 × 7 × 7)

$\Rightarrow \sqrt[3]{2744}\\[1em] \Rightarrow \sqrt[3]{(2 \times 2 \times 2) \times (7 \times 7 \times 7)}\\[1em] \Rightarrow 2 \times 7\\[1em] \Rightarrow 14$

Hence, Option 4 is the correct option.

Cube root of $1\dfrac{127}{216}$ is

1. $\dfrac{6}{7}$

2. $\dfrac{-7}{6}$

3. $1\dfrac{1}{6}$

4. $\dfrac{-6}{7}$

Answer

Converting mixed fraction to improper fraction:

$1\dfrac{127}{216} = \dfrac{343}{216}$

$\Rightarrow \sqrt[3]{1\dfrac{127}{216}}\\[1em] \Rightarrow \sqrt[3]{\dfrac{343}{216}}\\[1em] \Rightarrow \dfrac{\sqrt[3]{343}}{\sqrt[3]{216}}\\[1em] \Rightarrow \dfrac{\sqrt[3]{7 \times 7 \times 7}}{\sqrt[3]{(2 \times 2 \times 2) \times (3 \times 3 \times 3)}}\\[1em] \Rightarrow \dfrac{7}{2 \times 3}\\[1em] \Rightarrow \dfrac{7}{6}\\[1em] \Rightarrow 1\dfrac{1}{6}$

Hence, Option 3 is the correct option.

The smallest number by which 192 should be multiplied to make it a perfect cube is

1. 9

2. 6

3. 3

4. 2

Answer

Finding prime factors of 192 :

$\begin{array}{l|r} 2 & 192 \\ \hline 2 & 96 \\ \hline 2 & 48 \\ \hline 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$

192 = (2 × 2 × 2) × (2 × 2 × 2) × 3

Since 3 is left after grouping the prime factors in triplets, so two more 3's are needed to complete the triplet.

∴ The required number = 3 × 3 = 9

Hence, Option 1 is the correct option.

The smallest number by which 686 should be divided to make it a perfect cube is

1. 1

2. 2

3. 3

4. 4

Answer

Finding prime factors of 686 :

$\begin{array}{l|r} 2 & 686 \\ \hline 7 & 343 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$

686 = 2 × (7 × 7 × 7)

Since 2 is left after grouping the prime factors in triplets, so 686 should be divided by 2.

686 ÷ 2 = 343 = (7 × 7 × 7), which is a perfect cube.

Hence, Option 2 is the correct option.

The volume of a cube is 729 m³. Length of its side is

1. 3 m

2. 6 m

3. 9 m

4. 27 m

Answer

Let the length of the side of the cube be $a$ m.

We know that, volume of a cube = (side)³

Finding prime factors of 729 :

$\begin{array}{l|r} 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$

Expressing 729 into prime factors, we have

729 = (3 × 3 × 3) × (3 × 3 × 3)

$\Rightarrow a^3 = 729\\[1em] \Rightarrow a = \sqrt[3]{729}\\[1em] \Rightarrow a = \sqrt[3]{(3 \times 3 \times 3) \times (3 \times 3 \times 3)}\\[1em] \Rightarrow a = 3 \times 3\\[1em] \Rightarrow a = 9$

Hence, Option 3 is the correct option.