(B) Let the fraction be $x$. According to the problem,we multiply the fraction by itself and divide by its reciprocal: $x^2 \div (1/x) = 18 \frac{26}{

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(B) Let the fraction be $x$. According to the problem,we multiply the fraction by itself and divide by its reciprocal: $x^2 \div (1/x) = 18 \frac{26}{27}$ $x^2 \times x = \frac{18 \times 27 + 26}{27}$ $x^3 = \frac{486 + 26}{27} = \frac{512}{27}$ $x^3 = (\frac{8}{3})^3$ Taking the cube root on both sides,we get: $x = \frac{8}{3} = 2 \frac{2}{3}$

Competitive Quantitative Aptitude Number system Simplification

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If we multiply a fraction by itself and divide the product by its reciprocal,the fraction thus obtained is $18 \frac{26}{27}$. The fraction is

A
$\frac{8}{27}$
B
$2 \frac{2}{3}$
C
$1 \frac{1}{3}$
D
None of these

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What approximate value will come in place of the question mark $(?)$ in the given equation? (You are not expected to calculate the exact value)
$2775 \times \frac{160}{\sqrt{?}} = 5550$

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On simplification of $\frac{(2.644)^{2}-(2.356)^{2}}{0.288},$ we get

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Simplify: $2 \div [2 + 2 \div \{2 + 2 \div (2 + 2 \div 3)\}]$

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Simplify: $1-[2-\{5-(4-3-2)\}]$

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If the cube root of $79507$ is $43$,then the value of $\sqrt[3]{79.507} + \sqrt[3]{0.079507} + \sqrt[3]{0.000079507}$ is

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If we multiply a fraction by itself and divide the product by its reciprocal,the fraction thus obtained is $18 \frac{26}{27}$. The fraction is

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What approximate value will come in place of the question mark $(?)$ in the given equation? (You are not expected to calculate the exact value)$2775 \times \frac{160}{\sqrt{?}} = 5550$

On simplification of $\frac{(2.644)^{2}-(2.356)^{2}}{0.288},$ we get

Simplify: $2 \div [2 + 2 \div \{2 + 2 \div (2 + 2 \div 3)\}]$

Simplify: $1-[2-\{5-(4-3-2)\}]$

If the cube root of $79507$ is $43$,then the value of $\sqrt[3]{79.507} + \sqrt[3]{0.079507} + \sqrt[3]{0.000079507}$ is

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What approximate value will come in place of the question mark $(?)$ in the given equation? (You are not expected to calculate the exact value)
$2775 \times \frac{160}{\sqrt{?}} = 5550$