RS Aggarwal solution class 8 chapter 4 Cubes and Cube roots Test Paper 4

Test Paper 4

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Q1 Test Paper 4 Class 8 RS AGGARWAL chapter 4 Cubes and Cube roots

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Question 1:

Evaluate ${\left(1\frac{2}{5}\right)}^3.$

Answer 1:

${\left(1\frac{2}{5}\right)}^3$
${\left(1\frac{2}{5}\right)}^3$ = ${\left({\displaystyle \frac{7}{5}}\right)}^3 = \frac{{\left(7\right)}^3}{{\left(5\right)}^3} = \frac{\left(7\times 7\times 7\right)}{\left(5\times 5\times 5\right)} =\frac{343}{125}$
∴ ${\left(1\frac{2}{5}\right)}^3$ = $\frac{343}{125}$

Q2 Test Paper 4 Class 8 RS AGGARWAL chapter 4 Cubes and Cube roots

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Question 2:

Evaluate $\sqrt[3]{4096}.$

Answer 2:

$\sqrt[3]{4096}$

By prime factorisation method:

$\sqrt[3]{4096}$ = $\sqrt[3]{2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2} = \sqrt[3]{{\left(2\right)}^3\times {\left(2\right)}^3\times {\left(2\right)}^3\times {\left(2\right)}^3}$
$\sqrt[3]{4096}$ = $\left(2\right)\times \left(2\right)\times \left(2\right)\times \left(2\right) =16$.

∴ $\sqrt[3]{4096}$ = $16$

Q3 Test Paper 4 Class 8 RS AGGARWAL chapter 4 Cubes and Cube roots

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Question 3:

Evaluate $\sqrt[3]{216\times 343}.$

Answer 3:

$\sqrt[3]{216\times 343}$
By prime factorisation:

$\sqrt[3]{216\times 343}$ = $\sqrt[3]{216}\times \sqrt[3]{343} = \sqrt[3]{2\times 2\times 2\times 3\times 3\times 3}\times \sqrt[3]{7\times 7\times 7} = \sqrt[3]{{\left(2\right)}^3\times {\left(3\right)}^3}\times \sqrt[3]{{\left(7\right)}^3}$
$\sqrt[3]{216\times 343}$ = $\left(2\right)\times \left(3\right)\times \left(7\right) = 42$

∴ $\sqrt[3]{216\times 343}$ = $42$

Q4 Test Paper 4 Class 8 RS AGGARWAL chapter 4 Cubes and Cube roots

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Question 4:

Evaluate $\sqrt[3]{\frac{-64}{125}}.$

Answer 4:

$\sqrt[3]{\frac{-64}{125}}$
By prime factorisation method:

$\sqrt[3]{\frac{-64}{125}}$ = $\frac{\sqrt[3]{-64}}{\sqrt[3]{125}}= \frac{\sqrt[3]{\left(-4\right)\times \left(-4\right)\times \left(-4\right)}}{\sqrt[3]{5\times 5\times 5}} = \frac{\sqrt[3]{{\left(-4\right)}^3}}{\sqrt[3]{{\left(5\right)}^3}}$
$\sqrt[3]{\frac{-64}{125}}$ = $\frac{-4}{5}$
∴ $\sqrt[3]{\frac{-64}{125}}$ = $\frac{-4}{5}$

Q5 Test Paper 4 Class 8 RS AGGARWAL chapter 4 Cubes and Cube roots

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Question 5:

Mark (✓) against the correct answer
${\left(1\frac{3}{4}\right)}^3=?$
(a) $1\frac{27}{64}$
(b) $2\frac{27}{64}$
(c) $5\frac{23}{64}$
(d) none of these

Answer 5:

(c) $5\frac{23}{64}$
${\left(1\frac{3}{4}\right)}^3$ = ${\left(\frac{7}{4}\right)}^3 =\frac{{\left(7\right)}^3}{{\left(4\right)}^3} =\frac{7\times 7\times 7}{4\times 4\times 4}= \frac{343}{64}$
${\left(1\frac{3}{4}\right)}^3$ = $\frac{343}{64}$ = $5\frac{23}{64}$
∴ ${\left(1\frac{3}{4}\right)}^3$ = $5\frac{23}{64}$

Q6 Test Paper 4 Class 8 RS AGGARWAL chapter 4 Cubes and Cube roots

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Question 6:

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Which of the following numbers is a perfect cube?
(a) 121
(b) 169
(c) 196
(d) 216

Answer 6:

(d) 216



216 = $2\times 2\times 2\times 3\times 3\times 3 = {\left(2\right)}^3\times {\left(3\right)}^3 = {\left(6\right)}^3$

216 = 6³
Hence, 216 is a perfect cube.

Q7 Test Paper 4 Class 8 RS AGGARWAL chapter 4 Cubes and Cube roots

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Question 7:

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$\sqrt[3]{216\times 64}=?$
(a) 64
(b) 32
(c) 24
(d) 36

Answer 7:

(c) 24

$\sqrt[3]{216\times 64}$ = $\sqrt[3]{216}\times \sqrt[3]{64} = \sqrt[3]{2\times 2\times 2\times 3\times 3\times 3}\times \sqrt[3]{2\times 2\times 2\times 2\times 2\times 2}$
$\sqrt[3]{216\times 64}$ = $\sqrt[3]{{\left(2\right)}^3\times {\left(3\right)}^3}\times \sqrt[3]{{\left(2\right)}^3\times {\left(2\right)}^3} = \sqrt[3]{{\left(6\right)}^3}\times \sqrt[3]{{\left(4\right)}^3}$
$\sqrt[3]{216\times 64}$ = $6\times 4 = 24$
∴ $\sqrt[3]{216\times 64}$ =  24

Q8 Test Paper 4 Class 8 RS AGGARWAL chapter 4 Cubes and Cube roots

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Question 8:

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$\sqrt[3]{\frac{-343}{729}}=?$
(a) $\frac{7}{9}$
(b) $\frac{-7}{9}$
(c) $\frac{-9}{7}$
(d) $\frac{9}{7}$

Answer 8:

(b) $\frac{-7}{9}$
By prime factorisation:
$\sqrt[3]{\frac{-343}{729}}$ = $\frac{\sqrt[3]{-343}}{\sqrt[3]{729}} = \frac{\sqrt[3]{\left(-7\right)\times \left(-7\right)\times \left(-7\right)}}{\sqrt[3]{3\times 3\times 3\times 3\times 3\times 3}} = \frac{\sqrt[3]{{\left(-7\right)}^3}}{\sqrt[3]{{\left(3\right)}^3\times {\left(3\right)}^3}}$
$\sqrt[3]{\frac{-343}{729}}$ = $\frac{\sqrt[3]{{\left(-7\right)}^3}}{\sqrt[3]{{\left(9\right)}^3}}$ = $\frac{-7}{9}$

∴ $\sqrt[3]{\frac{-343}{729}}$ = $\frac{-7}{9}$

Q9 Test Paper 4 Class 8 RS AGGARWAL chapter 4 Cubes and Cube roots

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Question 9:

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By what least number should 324 be multiplied to get a perfect cube?
(a) 12
(b) 14
(c) 16
(d) 18

Answer 9:

(d) 18



324 = $2\times 2\times 3\times 3\times 3\times 3 = 2\times 2\times 3\times {\left(3\right)}^3$

Therefore, to show that the given number is the product of three triplets, we need to multiply 324 by $\left(2\times 3\times 3\right)$.
In other words, we need to multiply 324 by 18 to make it a perfect cube.

Q10 Test Paper 4 Class 8 RS AGGARWAL chapter 4 Cubes and Cube roots

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Question 10:

Mark (✓) against the correct answer
$\frac{\sqrt[3]{128}}{\sqrt[3]{250}}=?$
(a) $\frac{3}{5}$
(b) $\frac{4}{5}$
(c) $\frac{2}{5}$
(d) none of these

Answer 10:

(b) $\frac{4}{5}$

Resolving the numerator and the denominator into prime factors:

$$\begin{gathered} \frac{\sqrt[3]{128}}{\sqrt[3]{250}}=\sqrt[3]{\frac{128}{250}}=\sqrt[3]{\frac{2\times 8\times 8}{2\times 5\times 5\times 5}}=\sqrt[3]{\frac{\cancel{2}\times 8\times 8}{\cancel{2}\times 5\times 5\times 5}}=\sqrt[3]{\frac{8\times 8}{5\times 5\times 5}} \\ =\sqrt[3]{\frac{{\left(2\right)}^3\times {\left(2\right)}^3}{{\left(5\right)}^3}}=\frac{2\times 2}{5}=\frac{4}{5} \end{gathered}$$

Q11 Test Paper 4 Class 8 RS AGGARWAL chapter 4 Cubes and Cube roots

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Question 11:

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Which of the following is a cube of an odd number?
(a) 216
(b) 512
(c) 343
(d) 1000

Answer 11:

(c) 343
The cube of an odd number will always be an odd number.
Therefore, 343 is the cube of an odd number.

Q12 Test Paper 4 Class 8 RS AGGARWAL chapter 4 Cubes and Cube roots

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Question 12:

Fill in the blanks.
(i) $\sqrt[3]{ab}=\left(\sqrt[3]{a}\right)\times \left(.........\right).$
(ii) $\sqrt[3]{\frac{a}{b}}=.........$
(iii) $\sqrt[3]{-x}=.........$
(iv) (0.5)³ = .........

Answer 12:

(i) $\sqrt[3]{b}$

$\sqrt[3]{ab} = \left(\sqrt[3]{a}\right)\times \left(\sqrt[3]{b}\right)$

(ii) $\frac{\sqrt[3]{a}}{\sqrt[3]{b}}$

$\sqrt[3]{\frac{a}{b}} = \frac{\sqrt[3]{a}}{\sqrt[3]{b}}$

(iii) $-\sqrt[3]{x}$

$\sqrt[3]{-x} = -\sqrt[3]{x}$

(iv) 0.125

${\left(0.5\right)}^3 = \left(0.5\right)\times \left(0.5\right)\times \left(0.5\right) = 0.125$