RS Aggarwal solution class 8 chapter 4 Cubes and Cube roots Exercise 4D

Exercise 4D

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Q1 | Ex-4D | Class 8 | RS AGGARWAL | Cubes and Cube roots | Chapter 4 | myhelper

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Question 1:

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Which of the following numbers is a perfect cube?
(a) 141
(b) 294
(c) 216
(d) 496

Answer 1:

(a)
141 is not a perfect cube.

(b)
294 is not a perfect cube.

(c) (✓)
216 is a perfect cube.
216 = $\left(2\times 2\times 2\right)\times \left(3\times 3\times 3\right)=\left(2^3\right)\times \left(3^3\right)= 6^3$

(d)
496 is not a perfect cube.

Q2 | Ex-4D | Class 8 | RS AGGARWAL | Cubes and Cube roots | Chapter 4 | myhelper

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Question 2:

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Which of the following numbers is a perfect cube?
(a) 1152
(b) 1331
(c) 2016
(d) 739

Answer 2:

(a)
1152 = $2\times 2\times 2\times 2\times 2\times 2\times 2\times 3\times 3 = {\left(2\right)}^3\times {\left(2\right)}^3\times \left(2\times 3\times 3\right)$.
Hence, 1152 is not a perfect cube.

(b) (✓)
1331 = $11\times 11\times 11 = {\left(11\right)}^3$
Hence, 1331 is a perfect cube.

(c)
2016 = $2\times 2\times 2\times 2\times 2\times 3\times 3\times 7 = {\left(2\right)}^3\times 2\times 2\times 3\times 3\times 7$
Hence, 2016 is not a perfect cube.

(d)
739 is not a perfect cube.

Q3 | Ex-4D | Class 8 | RS AGGARWAL | Cubes and Cube roots | Chapter 4 | myhelper

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Question 3:

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$\sqrt[3]{512}=?$
(a) 6
(b) 7
(c) 8
(d) 9

Answer 3:

(c) 8

$\sqrt[3]{512}$ = $\sqrt[3]{2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2 } =\sqrt[3]{\left(2\times 2\times 2\right)\times \left(2\times 2\times 2\right)\times \left(2\times 2\times 2\right)}$
$\sqrt[3]{512}$ = $\sqrt[3]{{\left(2\right)}^3\times {\left(2\right)}^3\times {\left(2\right)}^3 }= 8$

Hence, the cube root of 512 is 8.

Q4 | Ex-4D | Class 8 | RS AGGARWAL | Cubes and Cube roots | Chapter 4 | myhelper

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Question 4:

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$\sqrt[3]{125\times 64}=?$
(a) 100
(b) 40
(c) 20
(d) 30

Answer 4:

(c) 20

$\sqrt[3]{125\times 64} = \sqrt[3]{125}\times \sqrt[3]{64} =\sqrt[3]{5\times 5\times 5}\times \sqrt[3]{2\times 2\times 2\times 2\times 2\times 2}$
$\sqrt[3]{125\times 64} = \sqrt[3]{{\left(5\right)}^3}\times \sqrt[3]{{\left(2\right)}^3\times {\left(2\right)}^3} = \sqrt[3]{{\left(5\right)}^3}\times \sqrt[3]{{\left(4\right)}^3}$
$\sqrt[3]{125\times 64} = 5\times 4 = 20$

Hence, the cube root of $\sqrt[3]{125\times 64}$ is 20.

Q5 | Ex-4D | Class 8 | RS AGGARWAL | Cubes and Cube roots | Chapter 4 | myhelper

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Question 5:

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$\sqrt[3]{\frac{64}{343}}=?$
(a) $\frac{4}{9}$
(b) $\frac{4}{7}$
(c) $\frac{8}{7}$
(d) $\frac{8}{21}$

Answer 5:

(b) $\frac{4}{7}$
$\sqrt[3]{\frac{64}{343}}$ = $\frac{\sqrt[3]{64}}{\sqrt[3]{343}} = \frac{\sqrt[3]{4\times 4\times 4}}{\sqrt[3]{7\times 7\times 7}}= \frac{\sqrt[3]{{\left(4\right)}^3}}{\sqrt[3]{{\left(7\right)}^3}}$
$\sqrt[3]{\frac{64}{343}}$ = $\frac{4}{7}$
∴ $\sqrt[3]{\frac{64}{343}}$ = $\frac{4}{7}$

Q6 | Ex-4D | Class 8 | RS AGGARWAL | Cubes and Cube roots | Chapter 4 | myhelper

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Question 6:

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$\sqrt[3]{\frac{-512}{729}}=?$
(a) $\frac{-7}{9}$
(b) $\frac{-8}{9}$
(c) $\frac{7}{9}$
(d) $\frac{8}{9}$

Answer 6:

(b) $\frac{-8}{9}$
$\sqrt[3]{\frac{-512}{729}}$= $\frac{\sqrt[3]{-512}}{\sqrt[3]{729}} = \frac{\sqrt[3]{\left(-8\right)\times \left(-8\right)\times \left(-8\right)}}{\sqrt[3]{9\times 9\times 9}} = \frac{\sqrt[3]{{\left(-8\right)}^3}}{\sqrt[3]{{\left(9\right)}^3}}$
$\sqrt[3]{\frac{-512}{729}}$ = $\frac{-8}{9}$
∴ $\sqrt[3]{\frac{-512}{729}}$ = $\frac{-8}{9}$

Q7 | Ex-4D | Class 8 | RS AGGARWAL | Cubes and Cube roots | Chapter 4 | myhelper

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Question 7:

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By what least number should 648 be multiplied to get a perfect cube?
(a) 3
(b) 6
(c) 9
(d) 8

Answer 7:

(c) 9

648 = $2\times 2\times 2\times 3\times 3\times 3\times 3 = {\left(2\right)}^3\times {\left(3\right)}^3\times 3$
Therefore, to get a perfect cube, we need to multiply 648 by 9, i.e. $\left(3\times 3\right)$.

Q8 | Ex-4D | Class 8 | RS AGGARWAL | Cubes and Cube roots | Chapter 4 | myhelper

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Question 8:

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By what least number should 1536 be divided to get a perfect cube?
(a) 3
(b) 4
(c) 6
(d) 8

Answer 8:

(a) 3


1536 = $2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 3 = {\left(2\right)}^3\times {\left(2\right)}^3\times {\left(2\right)}^3\times 3$
Therefore, to get a perfect cube, we need to divide 1536 by 3.

Q9 | Ex-4D | Class 8 | RS AGGARWAL | Cubes and Cube roots | Chapter 4 | myhelper

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Question 9:

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${\left(1\frac{3}{10}\right)}^3=?$
(a) $1\frac{27}{1000}$
(b) $2\frac{27}{1000}$
(c) $2\frac{197}{1000}$
(d) none of these

Answer 9:

(c) $2\frac{197}{1000}$
$$\begin{gathered} {\left(1\frac{3}{10}\right)}^3 = {\left({\displaystyle \frac{13}{10}}\right)}^3 = \frac{{\left(13\right)}^3}{{\left(10\right)}^3} = \frac{\left(13\times 13\times 13\right)}{\left(10\times 10\times 10\right)} \\ {\left(1\frac{3}{10}\right)}^3 = \frac{2197}{1000} = 2\frac{197}{1000} \end{gathered}$$

∴ ${\left(1\frac{3}{10}\right)}^3$ = $2\frac{197}{1000}$

Q10 | Ex-4D | Class 8 | RS AGGARWAL | Cubes and Cube roots | Chapter 4 | myhelper

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Question 10:

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(0.8)³ = ?
(a) 51.2
(b) 5.12
(c) 0.512
(d) none of these

Answer 10:

(c) 0.512

${\left(0.8\right)}^{3 } = \left(0.8\right)\times \left(0.8\right)\times \left(0.8\right) = 0.512$
∴ ${\left(0.8\right)}^{3 }$ = $0.512$