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Fundamental Laws of Logarithms

If x, y is ...

Question

If $(x, y)$ is a solution of $x^{2} + y^{2} = 160$ and $y = 3 x$ , then $y^{2}$ is equal to

12

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16

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120

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144

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Solution

The correct option is C $144$
Let $x^{2} + y^{2} = 160$ ---(1)
We have $y = 3 x$
Substitute the value of $y$ in the above equation, we get
$x^{2} + (3 x)^{2} = 160$
$∴ 10 x^{2} = 160$
$∴ x^{2} = 16$
Plug the value of $x^{2}$ in equation (1), we get:
$16 + y^{2} = 160$
$∴ y^{2} = 160 - 16$
$∴ y^{2} = 144$

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