2 Mark Question

Question 12 Marks

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

Answer

$\sqrt[3]{4096}$ By prime factorisation, we have$4096=\underline{2\times2\times2}\times\underline{2\times2\times2}\times\underline{2\times2\times2}$
$= (2\times2\times2)^3$ $=8^3$ $\therefore\sqrt[3]{4096}=8$

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Question 22 Marks

Evaluate:
$\sqrt[3]{-216}$

Answer

We know that $\sqrt[3]{-216}=-\sqrt[3]{216}$
Resolving 216 into prime factors,
We get,
$216= \underline{2\times2\times 2}\times\underline{3\times3\times3}$
$= (2\times 3)^3$
$=(6)^3$
$\therefore\sqrt[3]{216}=6 $
$\therefore\sqrt[3]{-216}=-(\sqrt[3]{216})=-6 $

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Question 32 Marks

Evaluate:
$\sqrt[3]{\frac{729}{1000}}$

Answer

$\sqrt[3]{\frac{729}{1000}}$$=\frac{\sqrt[3]{729}}{\sqrt[3]{1000}}$
$=\frac{\sqrt[3]{9\times9\times9}}{\sqrt[3]{10\times10\times10}}$
$=\frac{9}{10}$

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Question 42 Marks

Evaluate:$\Big(\frac{1}{15}\Big)^3$

Answer

$\Big(\frac{1}{15}\Big)^3$$=\Big(\frac{1}{15}\times\frac{1}{15}\times\frac{1}{15}\Big)$
$=\Big(\frac{1}{3375}\Big)$
Thus, the cube of $\Big(\frac{1}{15}\Big)$ is $\Big(\frac{1}{3375}\Big)$

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Question 52 Marks

Evaluate:
$\sqrt[3]{9261}$

Answer

$\sqrt[3]{9261}$ By prime factorisation, we have$9261=\underline{3\times3\times3}\times\underline{7\times7\times7}$
$= (3\times7)^3$ $= (21)^3$ $\therefore\sqrt[3]{9261}=21$

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Question 62 Marks

Evaluate:
$\sqrt[3]{-1331}$

Answer

We know that $\sqrt[3]{-1331}=-\sqrt[3]{1331}$
Resolving 1000 into prime factors,
We get,
$1331= 11\times11\times 11$
$= (11)^3$
$\therefore\sqrt[3]{1331}=11$
$\therefore\sqrt[3]{-1331}=-\big(\sqrt[3]{1331}\big)=-11$

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Question 72 Marks

Find the smallest number by which 1323 must be multiplied so that the product is a perfect cube.

Answer

$\begin{array}{c|c}3&1323\\\hline3&441\\\hline3&147\\\hline7&49\\\hline7&7\\\hline&1\end{array}$1323 = 3 × 3 × 3 × 7 × 7
To make it a perfect cube, it has to be multiplied by 7.

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Question 82 Marks

Which of the following numbers are perfect cubes? In case of perfect cube, find the number whose cube is the given number.
$3375$

Answer

$3375$
Resolving 3375 into prime factors:
$3375=3 \times 3 \times 3 \times 5 \times 5 \times 5$
Here, two triplet are formed, which are $3^3$ and $5^3$.
Hence, 3375 can be expressed as the product of the triplets of 3 and 5, i.e. $3^3 \times 5^3=15^3$.
Therefore, 3375 is a perfect cube.

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Question 92 Marks

Evaluate:
$\sqrt[3]{\frac{-512}{343}}$

Answer

$\sqrt[3]{\frac{-512}{343}}$$=\frac{\sqrt[3]{-512}}{\sqrt[3]{343}}$
$=\frac{\sqrt[3]{(-8)\times(-8)\times(-8)}}{\sqrt[3]{7\times7\times7}}$
$=\frac{-8}{7}$

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Question 102 Marks

Evaluate:
$\sqrt[3]{729}$

Answer

$\sqrt[3]{729}$ By prime factorisation, we have$729=\underline{3\times3\times3}\times\underline{3\times3\times3}$
$= (3\times3)^3$ $= 9^3$ $\therefore\sqrt[3]{729}=9$

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Question 112 Marks

Which of the following numbers are perfect cubes? In case of perfect cube, find the number whose cube is the given number.
$9261$

Answer

$9261$
Resolving 9261 into prime factors:
$9261=3 \times 3 \times 3 \times 7 \times 7 \times 7$
Here, two triplet are formed, which are $3^3$ and $7^3$.
Hence, 9261 can be expressed as the product of the triplets of 3 and 7 , i.e. $3^3 \times 7^3=21^3$.
Therefore, 9261 is a perfect cube.

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Question 122 Marks

Evaluate:
$\sqrt[3]{343}$

Answer

$\sqrt[3]{343}$ By prime factorisation, we have$343 = 7\times7\times7$
$= 7^3$ $\therefore\sqrt[3]{343}=7$

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Question 132 Marks

Evaluate:
$\sqrt[3]{8000}$

Answer

$\sqrt[3]{8000}$ By prime factorisation, we have$8000=\underline{2\times2\times2}\times\underline{2\times2\times2}\times\underline{5\times5\times5}$
$= (2\times2\times5)^3$ $=(20)^3$ $\therefore\sqrt[3]{8000}=20$

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Question 142 Marks

Evaluate:
$\sqrt[3]{64}$

Answer

$\sqrt[3]{64}$ By prime factorisation: $64 = \underline{2\times2\times 2}\times\underline{2\times2\times2}$ $= (2\times 2)^3$$= 4^3$
$\therefore\sqrt[3]{64}=4$

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Question 152 Marks

Evaluate:
$\sqrt[3]{64\times729}$

Answer

$\sqrt[3]{64\times729}$$=\sqrt[3]{64}\times\sqrt[3]{729}$
$=\sqrt[3]{4\times4\times4}\times\sqrt[3]{9\times9\times9}$
$=(4\times9)$
$=36$

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Question 162 Marks

Evaluate:
$\sqrt[3]{1728}$

Answer

$\sqrt[3]{1728}$ By prime factorisation, we have$1728=\underline{2\times2\times2}\times\underline{2\times2\times2}\times\underline{3\times3\times3}$
$= (2\times2\times3)^3$ $= (12)^3$ $\therefore\sqrt[3]{1728}=12$

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Question 172 Marks

Evaluate:
$\sqrt[3]{3375}$

Answer

$\sqrt[3]{3375}$ By prime factorisation, we have$3375=\underline{3\times3\times3}\times\underline{5\times5\times5}$
$= (3\times5)^3$ $=(15)^3$ $\therefore\sqrt[3]{3375}=15$

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Question 182 Marks

Evaluate:
$\sqrt[3]{\frac{125}{216}}$

Answer

$\sqrt[3]{\frac{125}{216}}$
$=\frac{\sqrt[3]{125}}{\sqrt[3]{216}}$
$=\frac{\sqrt[3]{5\times5\times5}}{\sqrt[3]{2\times2\times2\times3\times3\times3}}$
$=\frac{5}{2\times3}$
$=\frac{5}{6}$

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Question 192 Marks

Which of the following numbers are perfect cubes? In case of perfect cube, find the number whose cube is the given number.
8000

Answer

8000
Resolving 8000 into prime factors:
$8000=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 5$
Here, three triplet are formed, which are $2^3, 2^3$ and $5^3$.
Hence, 8000 can be expressed as the product of the triplets of 2,2 and 5, i.e. $2^3 \times 2^3 \times 5^3=20^3$.
Therefore, 8000 is a perfect cube.

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Question 202 Marks

Evaluate:
$\sqrt[3]{\frac{-27}{125}}$

Answer

$\sqrt[3]{\frac{-27}{125}}$
$=\frac{\sqrt[3]{-27}}{\sqrt[3]{125}}$
$=\frac{\sqrt[3]{(-3)\times(-3)\times(-3)}}{\sqrt[3]{5\times5\times5}}$
$=\frac{-3}{5}$

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Question 212 Marks

Evaluate:$\Big(\frac{4}{7}\Big)^3$

Answer

$\Big(\frac{4}{7}\Big)^3$$=\Big(\frac{4}{7}\times\frac{4}{7}\times\frac{4}{7}\Big)$
$=\Big(\frac{64}{343}\Big)$
Thus, the cube of $\Big(\frac{4}{7}\Big)$ is $\Big(\frac{64}{343}\Big)$

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Question 222 Marks

Which of the following numbers are perfect cubes? In case of perfect cube, find the number whose cube is the given number.
343

Answer

343
Resolving 343 into prime factors:
$343=7 \times 7 \times 7$
Here, one triplet is formed, which is $7^3$.
Hence, 343 can be expressed as the product of the triplets of 7.
Therefore, 343 is a perfect cube.

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Question 232 Marks

Evaluate:
$\sqrt[3]{\frac{-64}{343}}$

Answer

$\sqrt[3]{\frac{-64}{343}}$
$=\frac{\sqrt[3]{-64}}{\sqrt[3]{343}}$
$=\frac{\sqrt[3]{(-4)\times(-4)\times(-4)}}{\sqrt[3]{7\times7\times7}}$
$=\frac{-4}{7}$

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Question 242 Marks

Which of the following numbers are perfect cubes? In case of perfect cube, find the number whose cube is the given number.
$125$

Answer

125
Resolving 125 into prime factors:
$125=5 \times 5 \times 5$
Here, one triplet is formed, which is $5^3$.
Hence, 125 can be expressed as the product of the triplets of 5 .
Therefore, 125 is a perfect cube.

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Question 252 Marks

Evaluate:$\Big(1\frac{3}{10}\Big)^3$

Answer

$\Big(1\frac{3}{10}\Big)^3$$=\Big(\frac{13}{10}\Big)^3$
$=\Big(\frac{13}{10}\times\frac{13}{10}\times\frac{13}{10}\Big)$
$=\Big(\frac{2197}{1000}\Big)$
Thus, the cube of $\Big(1\frac{3}{10}\Big)$ is $\Big(\frac{2197}{1000}\Big)$

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Question 262 Marks

Evaluate:
$\sqrt[3]{\frac{27}{64}}$

Answer

$\sqrt[3]{\frac{27}{64}}$
$=\frac{\sqrt[3]{27}}{\sqrt[3]{64}}$
$=\frac{\sqrt[3]{3\times3\times3}}{\sqrt[3]{4\times4\times4}}$
$=\frac{3}{4}$

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Question 272 Marks

Find the value of using the short-cut method:
$(25)^3$

Answer

$(25)^3$

We know that short-cut method for finding the cube of any two digit numbers is as given.

$(a+b)^3=a^3+3 a^2 b+3 a b^2+b^3$

$=a^2 \times a+a^2 \times 3 b+b^2 \times 3 a+b^2 \times b$

Here,

$a=2$

$b=5$

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Question 282 Marks

Evaluate:
$\sqrt[3]{-512}$

Answer

We know that $\sqrt[3]{-512}=-\sqrt[3]{512}$
Resolving 512 into prime factors,
We get,
$512= \underline{2\times2\times 2}\times\underline{2\times2\times2}\times\underline{2\times2\times2}$
$= (2\times2\times2)^3$
$=(8)^3$
$\therefore\sqrt[3]{512}=8$
$\therefore\sqrt[3]{-512}=-\big(\sqrt[3]{512}\big)=-8$

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Question 292 Marks

Evaluate:$\Big(\frac{10}{11}\Big)^3$

Answer

$\Big(\frac{10}{11}\Big)^3$$=\Big(\frac{10}{11}\times\frac{10}{11}\times\frac{10}{11}\Big)$
$=\Big(\frac{1000}{1331}\Big)$
Thus, the cube of $\Big(\frac{10}{11}\Big)$ is $\Big(\frac{1000}{1331}\Big)$

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