Compare $\frac{8}{16} □ \frac{8}{4}$

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None of these

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Solution

The correct option is B <
$\frac{8}{16} = \frac{1}{2} a n d \frac{8}{4} = 2$

obviously $\frac{1}{2} < 2$

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Similar questions

2 x + \frac{y}{3} = 12 and x y = 30, then 8 x^{3} + \frac{y^{3}}{27} =_______

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\frac{8}{12}

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\frac{8 - 4}{12 - 4}

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