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If xy > 0, is y/x + x/y > 2?
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03 Jul 2019, 07:00
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If \(xy > 0\), is \(\frac{y}{x} + \frac{x}{y} > 2\)? (1) \(y = x  1\) (2) \(x = 2y\)
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If xy > 0, is y/x + x/y > 2?
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03 Jul 2019, 07:10
\(\frac{x^2+y^2}{xy}>2\)
Since xy>0, we can multiply the stem by xy maintaining the same inequality and we know that neither x nor y can be 0
So, \(x^2+y^2>2xy\) or \(x^2+y^22xy>0\) or \((xy)^2>0\)
So we need to find whether \((xy)^2>0\), this will be true when xy is not equal to 0
(1) \(y=x1\)
So, \(xy=1\) => We know that xy is non zero
So Sufficient
(2) \(x=2y\)
Or, \(xy=y\)
Since y is non zero, \((xy)^2 > 0\)
Sufficient
Answer is (D)




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Re: If xy > 0, is y/x + x/y > 2?
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03 Jul 2019, 07:14
xy > 0 > Both x>0 & y>0 (OR) x<0 & y<0
(1) y = x  1 Since both x & y are of same sign, Irrespective of their values, x/y + y/x will always be > 2
eg: x = 2, y = 1 > 2/1 + 1/2 = 2.5 > 2
x = 5, y = 4 > 5/4 + 4/5 = 1.25 + 0.8 = 2.05 > 2
Sufficient
(2) x = 2y > x/y = 2 > x/y + y/x = 2 + 1/2 = 2.5 > 2
Sufficient
IMO Option D
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Re: If xy > 0, is y/x + x/y > 2?
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03 Jul 2019, 07:18
Question Stem \(\frac{y}{x}+\frac{x}{y}>2\) \(y^2+x^2>2xy\) \(y^2+x^22xy>0\) \((xy)^2>0\)
From Statement 1 \(xy=1\) > square both sides \((xy)^2=1\) sufficient
From Statement 2 \(x=2y\) \(x/y=2\) \(y/x=1/2\) \(\frac{x}{y}+\frac{y}{x}=2+\frac{1}{2}\)
Which is greater than 2, hence sufficient
Answer D



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Re: If xy > 0, is y/x + x/y > 2?
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03 Jul 2019, 07:18
Given, xy > 0 means neither x nor y can be ZERO



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Re: If xy > 0, is y/x + x/y > 2?
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03 Jul 2019, 07:28
Quote: If xy>0, is y/x+x/y>2?
(1) y=x−1
(2) x=2y If xy>0, i.e. signs of both x and y are identical Now, y/x+x/y will be equal to 2 only if both x and y are equal otherwise it will always be greater than 2? So question becomes Is x = y?(1) y=x−1 i.e. x and y are not equal hence SUFFICIENT (2) x=2y i.e. x and y are NOT equal hence SUFFICIENT Answer: Option D
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Re: If xy > 0, is y/x + x/y > 2?
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03 Jul 2019, 21:03
xy > 0, So x and y will have the same sign. We are required to find whether \(\frac{y}{x}\) + \(\frac{x}{y}\) > 2 => \(x^2 + y^2 > 2xy\) => \((x + y)^2 > 4xy\) > [a]
(1) y = x 1 => x = y + 1 Substituting the above value in [a] we get, \((y + 1 + y)^2 > 4 (y + 1) y\) => \(4y^2 + 4y + 1 > 4y^2 + 4y\) => \(1 > 0\) This will always be true. Hence Sufficient
(2) x = 2y Substituting the above value in [a] we get, \((2y + y)^2 > 4 (2y) y\) => \(9y^2 > 8y^2\) => \(9 > 8\) This will always be true. Hence sufficient
Answer D




Re: If xy > 0, is y/x + x/y > 2?
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03 Jul 2019, 21:03






