S.chand Books Class 8 Maths Solution Chapter Cube and Cube roots Exercise 4A

EXERCISE 4A

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Q1 | Ex-4A | Class 8 | Cube and Cube roots |S.Chand | Composite Mathematics | Chapter 4 | myhelper

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Question 1

Write (T) for True or (F) for false:

(i) The cube root of 8000 is 200.

Sol: False

(ii) Each prime factor appears 3 times in its cube.

Sol: True

(iii) ​$\sqrt[3]{27+64}=\sqrt[3]{27}+\sqrt[3]{64}$​.

Sol: False

(iv) For an integer a, ​$a^3$​ is always greater than ​$a^2$​.

Sol: False

(v) The least number by which 72 must be divided to make it a perfect cube is 9.

Sol: True

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Q2 | Ex-4A | Class 8 | Cube and Cube roots |S.Chand | Composite Mathematics | Chapter 4 | myhelper

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Find the cubes of the following numbers.

Question 2

(i) 8

Sol: 8×8×8

=512

(ii) -15

Sol: (-15)×(-15)×(-15)

= -3375

(iii) 600

Sol:600×600×600

= 216000000

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Q3 | Ex-4A | Class 8 | Cube and Cube roots |S.Chand | Composite Mathematics | Chapter 4 | myhelper

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Question 3

(i) ​ ​$\dfrac{5}{6}$​

Sol: 

Cube :

$=\dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{5}{6}=\dfrac{5\times5\times 5}{6\times 6\times 6}$

$=\dfrac{125}{216}$

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(ii) $\dfrac{-7}{9}$ 

Sol :

Cube :

$=\dfrac{-7}{9} \times \dfrac{-7}{9} \times \dfrac{-7}{9}=$

$=\frac{-7 \times -7 \times -7}{9 \times 9 \times 9}$

$=\frac{-343}{729}$

(iii) ​$1\dfrac{3}{5}$​

Sol :

$=\frac{1 \times 5+3}{5}=\frac{8}{5}$

Cube :

$=\dfrac{8}{5} \times \dfrac{8}{5} \times \dfrac{8}{5}$

$=\frac{8 \times 8 \times 8}{5 \times 5 \times 5}=\frac{512}{125}$

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Q4 | Ex-4A | Class 8 | Cube and Cube roots |S.Chand | Composite Mathematics | Chapter 4 | myhelper

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Question 4

(i) 0.03

Sol : 0.03×0.03×0.03

=0.000027

(ii) 1.7

Sol: 1.7×1.7×1.7

=4.913

(iii) ​ -0.008

Sol: -0.008×0.008×0.008

=0.000000512

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Q5 | Ex-4A | Class 8 | Cube and Cube roots |S.Chand | Composite Mathematics | Chapter 4 | myhelper

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Question 5

Which of the following numbers are perfect cubes?

65, 128, 243, 512, 900, 1728, 4096

Sol:

65=5×13 (Not a perfect cube triplet not found)

128=2×2×2×2×2×2×2 (triplet not found)

243=3×3×3×3×3 (triplet not found)

512=2×2×2×2×2×2×2×2×2 (perfect cube , triplet found )

900=2×2×3×3×5×5 (triplet not found)

1728=2×2×2×2×2×2×3×3×3 (perfect cube , triplet found )

4096=2×2×2×2×2×2×2×2×2×2×2×2 (perfect cube , triplet found )

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Q6 | Ex-4A | Class 8 | Cube and Cube roots |S.Chand | Composite Mathematics | Chapter 4 | myhelper

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Question 6

Find the cube root of the following numbers by prime factorization method.

(i) 5832

Sol:

$\begin{array}{c|l} 2 & 5832 \\ \hline 2 & 2916 \\ \hline 2 & 1458 \\ \hline 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

$=\sqrt[3]{2^{3} \times 3^{6}}$

$=\sqrt[3]{2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3}$

=2×3×3=18

(ii) 91125 

Sol:

$\begin{array}{l|l} 3 & 91125 \\ \hline 3 & 30375\\ \hline 3 & 10125 \\ \hline 3 & 3315 \\ \hline 3 & 1125 \\ \hline 3 & 375 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

$=\sqrt[3]{3^{6} \times 5^{3}}$

$=\sqrt[3]{3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 5 \times 5 \times 5}$

=3×5×5=45

(iii) ​ -9261

Sol:

$\begin{array}{l|l} 3 & 9261 \\ \hline 3 & 3087 \\ \hline 3 & 1029 \\ \hline 7 & 343 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

$=-\sqrt[3]{3^{3} \times 7^{3}}$

$=-\sqrt[3]{3 \times 3 \times 3 \times 7 \times 7 \times 7}$

=-(3×7)=-21

(iv) ​$\dfrac{125}{343}$​

Sol :

$\begin{array}{l|c} 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline &1 \end{array}$

$\begin{array}{c|c} 7 & 343 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline &1 \end{array}$

$=\sqrt[3]{\frac{5^{3}}{7^{3}}}$

$=\frac{5}{7}$

(v) ​$-\dfrac{2744}{4096}$​

Sol :

$\begin{array}{l|l} 2 & 2744 \\ \hline 2 & 1372 \\ \hline2 & 686 \\ \hline 7 & 343 \\ \hline 7& 49 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

$\begin{array}{c|c} 2 & 4096 \\ \hline 2 & 2048 \\ \hline 2 & 1024 \\ \hline 2 & 512 \\ \hline 2 & 256 \\ \hline 2 & 128 \\ \hline 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$

$=-\sqrt[3]{\frac{2^{3} \times 7^{3}}{2^{3}\times 2^{3} \times 2^{3} \times 2^{3}}}$

$=-\frac{7}{2\times 2 \times 2}=-\frac{7}{8}$

(vi) ​$-5\dfrac{104}{125}$​

Sol:

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

$\begin{array}{l|l} 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline &1 \end{array}$

$=-\sqrt[3]{\frac{5 \times 125+104}{125}}$

$=-\sqrt[3]{\frac{729}{125}}$

$=-\sqrt[3]{\frac{3 \times 3 \times 3 \times 3 \times 3 \times 3}{5 \times 5 \times 5}}$

$=-\frac{3 \times 3}{5}=-\frac{9}{5}=-1 \frac{4}{5}$

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Q7 | Ex-4A | Class 8 | Cube and Cube roots |S.Chand | Composite Mathematics | Chapter 4 | myhelper

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Question 7

Evaluate:

(i) ​$\sqrt[3]{1.331}$​

Sol :

$\begin{array}{c|c} \text { 11 } & 1331 \\ \hline 11&121 \\ \hline 11&11 \\ \hline &1 \end{array}$

$=\sqrt[3]{\frac{1331}{1000}}$

$=\sqrt[3]{\frac{11 \times 11 \times 11}{10 \times 10 \times 10}}$

$=\frac{11}{10}=1.1$

(ii) ​$\sqrt[3]{0.003375}$​

Sol :

$\begin{array}{c|c} 3 & 3315 \\ \hline 3 & 1125 \\ \hline 3 & 375 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline &1 \end{array}$

$=\sqrt[3]{\frac{3375}{1000000}}$

$=\sqrt[3]{\frac{3 \times 3 \times 3 \times 5 \times 5 \times 5}{10 \times 10 \times 10 \times 10 \times 10 \times 10}}$

$=\frac{3 \times 5}{10 \times 10}=\frac{15}{100}$

=0.15

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Q8 | Ex-4A | Class 8 | Cube and Cube roots |S.Chand | Composite Mathematics | Chapter 4 | myhelper

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Question 8

What is the smallest number by which each of the following numbers must be multiplied so that the product is a perfect cube.Also, find cube root of the product.

(i) 1125

Sol:

(ii) 6912

Sol:

(iii) 47916

Sol:

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Q9 | Ex-4A | Class 8 | Cube and Cube roots |S.Chand | Composite Mathematics | Chapter 4 | myhelper

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Question 9

Find the smallest number by which each of the following numbers must be divided so that quotient is a perfect cube. Also, find the cube root of the product.

(i) 3584

Sol:

(ii) 1458

Sol:

(iii) 120393

Sol:

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Q10 | Ex-4A | Class 8 | Cube and Cube roots |S.Chand | Composite Mathematics | Chapter 4 | myhelper

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Question 10

Find the value of :

(i) ​$\sqrt[3]{27}+\sqrt[3]{0.008}+\sqrt[3]{0.064}$​

Sol :

$\begin{array}{c|c} 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$
$\begin{array}{c|c} 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$
$\begin{array}{c|c} 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2&4 \\ \hline 2&2 \\ \hline & 1 \end{array}$

$=\sqrt[3]{3 \times 3 \times 3}+\sqrt[3]{\frac{8}{1000}}+\sqrt[3]{\frac{64}{1000}}$

$=3+\sqrt[3]{\frac{2 \times 2 \times 2}{10 \times 10 \times 10}}+\sqrt[3]{\frac{2 \times 2 \times 2 \times 2 \times 2 \times 2}{10 \times 10 \times 10}}$

$=3+\frac{2}{10}+\frac{2 \times 2}{10}$

$=\frac{3}{1}+\frac{2}{10}+\frac{4}{10}$

$=\frac{3\times 10+2\times 1+4 \times 1}{10}$

$=\frac{30+2+4}{10}=\frac{36}{10}$

=3.6

(ii) ​$\bigg\{(5^2+\sqrt{10^2})\bigg\}^3$​

Sol :

$\begin{array}{c|c} 2 & 100 \\ \hline 2 & 50 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

$=\{25+\sqrt{100}\}^{3}$

$=\{25+\sqrt{2 \times 2 \times 5 \times 5}\}^{3}$

$=\{25+(2 \times 5)\}^{3}$

$=(25+10)^{3}=(35)^{3}$

=35×35×35=42875

(iii) ​$\sqrt[3]{686}\times\sqrt[3]{500}$​

Sol :

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

$\begin{array}{c|c}2 & 500 \\ \hline 2 & 250 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

$=\sqrt[3]{2 \times 7 \times 7 \times 7} \times \sqrt[3]{2 \times 2 \times 5 \times 5 \times 5}$

$=7 \times \sqrt{2} \times \sqrt{2 \times 2} \times 5$

$=7 \times 5 \times \sqrt{2 \times 2 \times 2}$

=35×2=70