EXERCISE 4A
Q1 | Ex-4A | Class 8 | Cube and Cube roots |S.Chand | Composite Mathematics | Chapter 4 | myhelper
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
Q2 | Ex-4A | Class 8 | Cube and Cube roots |S.Chand | Composite Mathematics | Chapter 4 | myhelper
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
Q3 | Ex-4A | Class 8 | Cube and Cube roots |S.Chand | Composite Mathematics | Chapter 4 | myhelper
Question 3
(i) \( \dfrac{5}{6} \)
Sol:
Cube :
$=\dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{5}{6}=\dfrac{5\times5\times 5}{6\times 6\times 6}$
$=\dfrac{125}{216}$
(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 :
$=\dfrac{8}{5} \times \dfrac{8}{5} \times \dfrac{8}{5}$
$=\frac{8 \times 8 \times 8}{5 \times 5 \times 5}=\frac{512}{125}$
Q4 | Ex-4A | Class 8 | Cube and Cube roots |S.Chand | Composite Mathematics | Chapter 4 | myhelper
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
Q5 | Ex-4A | Class 8 | Cube and Cube roots |S.Chand | Composite Mathematics | Chapter 4 | myhelper
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 )
Q6 | Ex-4A | Class 8 | Cube and Cube roots |S.Chand | Composite Mathematics | Chapter 4 | myhelper
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:
$\begin{array}{l|l}3 & 729 \\ \hline 3 & 243 \\\hline 3 & 81 \\
\hline 3 & 27 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}$
$\begin{array}{l|l}
5 & 125 \\
\hline 5 & 25 \\
\hline 5 & 5 \\
\hline &1
\end{array}$
$=-\sqrt[3]{\frac{5 \times 125+104}{125}}$
$=-\sqrt[3]{\frac{729}{125}}$
$=-\sqrt[3]{\frac{3 \times 3 \times 3 \times 3 \times 3 \times 3}{5 \times 5 \times 5}}$
$=-\frac{3 \times 3}{5}=-\frac{9}{5}=-1 \frac{4}{5}$
Q7 | Ex-4A | Class 8 | Cube and Cube roots |S.Chand | Composite Mathematics | Chapter 4 | myhelper
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
Q8 | Ex-4A | Class 8 | Cube and Cube roots |S.Chand | Composite Mathematics | Chapter 4 | myhelper
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:
Q9 | Ex-4A | Class 8 | Cube and Cube roots |S.Chand | Composite Mathematics | Chapter 4 | myhelper
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:
Q10 | Ex-4A | Class 8 | Cube and Cube roots |S.Chand | Composite Mathematics | Chapter 4 | myhelper
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}$
$\begin{array}{c|c}
2 & 64 \\
\hline 2 & 32 \\
\hline 2 & 16 \\
\hline 2 & 8 \\
\hline 2&4 \\
\hline 2&2 \\
\hline & 1
\end{array}$
$=3+\frac{2}{10}+\frac{2 \times 2}{10}$
$=\frac{3}{1}+\frac{2}{10}+\frac{4}{10}$
$=\frac{3\times 10+2\times 1+4 \times 1}{10}$
$=\frac{30+2+4}{10}=\frac{36}{10}$
=3.6
(ii) \( \bigg\{(5^2+\sqrt{10^2})\bigg\}^3 \)
Sol :
$\begin{array}{c|c}2 & 100 \\
\hline 2 & 50 \\
\hline 5 & 25 \\
\hline 5 & 5 \\
\hline & 1
\end{array}$
$=\{25+\sqrt{100}\}^{3}$
$=\{25+\sqrt{2 \times 2 \times 5 \times 5}\}^{3}$
$=\{25+(2 \times 5)\}^{3}$
(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
Nice website 👍🏻
ReplyDelete