$\frac{\sqrt[3]{8}}{\sqrt{16}} \div \sqrt{\frac{100}{49}} \times \sqrt[3]{125} =$

$7$

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$1 \frac{3}{4}$

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$\frac{7}{100}$

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$\frac{4}{7}$

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Solution

The correct option is B
$1 \frac{3}{4}$

Explanation for the correct option:
Simplify the expression:
$\begin{array}{l} \frac{\sqrt[3]{8}}{\sqrt{16}} \div \frac{\sqrt{100}}{\sqrt{49}} \times \sqrt[3]{125} \\ = \frac{\sqrt[3]{2^{3}}}{\sqrt{4^{2}}} + \frac{\sqrt{10^{2}}}{\sqrt{7^{2}}} \times \sqrt[3]{5^{3}} \\ = \frac{{(2^{3})}^{\frac{1}{3}}}{{(4^{2})}^{\frac{1}{2}}} + \frac{{(10^{2})}^{\frac{1}{2}}}{{(7^{2})}^{\frac{1}{2}}} \times {(5^{3})}^{\frac{1}{3}} [∵ {(a^{m})}^{n} = a^{m \times n}] \\ = \frac{2}{4} \div \frac{10}{7} \times 5 \\ = \frac{2}{4} \times \frac{7}{10} \times 5 \\ = \frac{7}{4} \\ = 1 \frac{3}{4} \end{array}$
Hence, option (B) is the correct answer

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Similar questions

Q2. Find:

(i) (ii) (iii)

(iv)

Find $x$ , if:

(i) $4^{2 x} = \frac{1}{32}$

(ii) $\sqrt{2^{x + 3}} = 16$

(iii) ${(\sqrt{\frac{3}{5}})}^{x + 1} = \frac{125}{27}$

(iv) ${(\sqrt[3]{\frac{2}{3}})}^{x - 1} = \frac{27}{8}$

9^{\frac{3}{2}}

32^{\frac{2}{5}}

16^{\frac{3}{4}}

125^{\frac{- 1}{3}}

Simplify : $\frac{\sqrt[3]{8}}{\sqrt{16}} \div \frac{\sqrt{100}}{\sqrt{49}} \times \sqrt[3]{125}$

\frac{\sqrt[3]{8}}{\sqrt{16}} \div \frac{\sqrt{100}}{\sqrt{49}} \times \sqrt[3]{125}

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