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UPSC CDS 1 2022 Math Paper

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Question : 31 of 100

Marks: +1, -0

If

\frac{x-y}{x\sqrt{y}+y\sqrt{x}}=\frac{1}{\sqrt{x}}\quad(x>0,\;y>0)

then what is the value of

\frac{x}{y}

?

1
2
4
8

Solution: 👈: Video Solution

Given:

\frac{x-y}{x\sqrt{y}+y\sqrt{x}}=\frac{1}{\sqrt{x}}\quad(x>0,\;y>0)

Formula Used:

a^{2}-b^{2}=(a+b)(a-b)

Calculation:

We have

\frac{x-y}{x\sqrt{y}+y\sqrt{x}}=\frac{1}{\sqrt{x}}

\Rightarrow \frac{(\sqrt{x})^{2}-(\sqrt{y})^{2}}{\sqrt{x}\sqrt{x}\sqrt{y}+\sqrt{y}\sqrt{y}\sqrt{x}}=\frac{1}{\sqrt{x}}

\Rightarrow \frac{(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})}{\sqrt{x}\cdot\sqrt{y}(\sqrt{x}+\sqrt{y})}=\frac{1}{\sqrt{x}}

\Rightarrow \frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}\cdot\sqrt{y}}=\frac{1}{\sqrt{x}}

\Rightarrow \sqrt{x}-\sqrt{y}=\sqrt{y}

\Rightarrow \sqrt{x}=2\sqrt{y}

On squaring both side, we get

\Rightarrow x=4y

\Rightarrow \frac{x}{y}=4

\therefore

The required value of

\frac{x}{y}

is 4 .

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