In the following questions,two equations numb | vedclass

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(C) $I$. $\frac{5}{7} - \frac{5}{21} = \frac{\sqrt{x}}{42}$$\Rightarrow \frac{15 - 5}{21} = \frac{\sqrt{x}}{42}$$\Rightarrow \frac{10}{21} = \frac{\sq

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$x < y$

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(C) $I$. $\frac{5}{7} - \frac{5}{21} = \frac{\sqrt{x}}{42}$$\Rightarrow \frac{15 - 5}{21} = \frac{\sqrt{x}}{42}$$\Rightarrow \frac{10}{21} = \frac{\sqrt{x}}{42}$$\Rightarrow \sqrt{x} = \frac{10}{21} 	imes 42 = 20$$	herefore x = 20^2 = 400$$II$. $\frac{\sqrt{y}}{4} + \frac{\sqrt{y}}{16} = \frac{250}{\sqrt{y}}$$\Rightarrow \frac{4\sqrt{y} + \sqrt{y}}{16} = \frac{250}{\sqrt{y}}$$\Rightarrow \frac{5\sqrt{y}}{16} = \frac{250}{\sqrt{y}}$$\Rightarrow 5(\sqrt{y})^2 = 250 	imes 16$$\Rightarrow 5y = 4000$$\Rightarrow y = \frac{4000}{5} = 800$Comparing the values,$800 > 400$,therefore $y > x$ or $x < y$.

In the following questions,two equations numbered $I$ and $II$ are given. You have to solve both the equations and give the answer.
$I$: $\frac{5}{7} - \frac{5}{21} = \frac{\sqrt{x}}{42}$
$II$: $\frac{\sqrt{y}}{4} + \frac{\sqrt{y}}{16} = \frac{250}{\sqrt{y}}$

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In the following questions,two equations numbered $I$ and $II$ are given. You have to solve both the equations and give the answer.$I$: $\frac{5}{7} - \frac{5}{21} = \frac{\sqrt{x}}{42}$$II$: $\frac{\sqrt{y}}{4} + \frac{\sqrt{y}}{16} = \frac{250}{\sqrt{y}}$