I- -289 x -25-0II- -676 y -10-0A- x-yB- x - yC- x - yD- x - yE- x - y or the relationship cannot be established

The correct option is A x gt; yEquation I √(289)x + √(25) = 0 ⇒ 17x + 5 = 0 = 5/17 Equation II √(676)y + 10 = 0 ⇒ 26y + 10 = 0 ⇒ 26y + 10 = 0 ⇒ y = 10/26 ...

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Quantitative Aptitude

Quadratic Equations

I. √289 x +√2...

Question

I. $\sqrt{289} x + \sqrt{25} = 0$
II. $\sqrt{676} y + 10 = 0$

x > y

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x \geq y

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x < y

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x \leq y

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x = y

or the relationship cannot be established

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Solution

The correct option is A $x > y$
Equation I
$\sqrt{289} x + \sqrt{25} = 0$
$\Rightarrow 17 x + 5 = 0 = \frac{- 5}{17}$
Equation II
$\sqrt{676} y + 10 = 0$
$\Rightarrow 26 y + 10 = 0$
$\Rightarrow 26 y + 10 = 0$
$\Rightarrow y = \frac{- 10}{26}$
$\Rightarrow y = \frac{- 5}{13}$
Hence, x > y

Suggest Corrections

I. √289x+√25=0 II. √676y+10=0

The correct option is A x>yEquation I √289x+√25=0 ⇒17x+5=0=−517 Equation II √676y+10=0 ⇒26y+10=0 ⇒26y+10=0 ⇒y=−1026 ⇒y=−513 Hence, x > y

I. $\sqrt{289} x + \sqrt{25} = 0$
II. $\sqrt{676} y + 10 = 0$

The correct option is A $x > y$
Equation I
$\sqrt{289} x + \sqrt{25} = 0$
$\Rightarrow 17 x + 5 = 0 = \frac{- 5}{17}$
Equation II
$\sqrt{676} y + 10 = 0$
$\Rightarrow 26 y + 10 = 0$
$\Rightarrow 26 y + 10 = 0$
$\Rightarrow y = \frac{- 10}{26}$
$\Rightarrow y = \frac{- 5}{13}$
Hence, x > y