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If xy > 0, is y/x + x/y > 2?

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If xy > 0, is y/x + x/y > 2?  [#permalink]

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New post 03 Jul 2019, 07:00
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D
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Question Stats:

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Director
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If xy > 0, is y/x + x/y > 2?  [#permalink]

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New post 03 Jul 2019, 07:10
8
\(\frac{x^2+y^2}{xy}>2\)

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 \(x^2+y^2-2xy>0\) or \((x-y)^2>0\)

So we need to find whether \((x-y)^2>0\), this will be true when x-y is not equal to 0

(1) \(y=x-1\)

So, \(x-y=1\) => We know that x-y is non zero

So Sufficient

(2) \(x=2y\)

Or, \(x-y=y\)

Since y is non zero, \((x-y)^2 > 0\)

Sufficient

Answer is (D)
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Re: If xy > 0, is y/x + x/y > 2?  [#permalink]

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New post 03 Jul 2019, 07:14
2
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?  [#permalink]

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New post 03 Jul 2019, 07:18
2
Question Stem
\(\frac{y}{x}+\frac{x}{y}>2\)
\(y^2+x^2>2xy\)
\(y^2+x^2-2xy>0\)
\((x-y)^2>0\)

From Statement 1 \(x-y=1\) -> square both sides
\((x-y)^2=1\) 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?  [#permalink]

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New post 03 Jul 2019, 07:18
2
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?  [#permalink]

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New post 03 Jul 2019, 07:28
1
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?  [#permalink]

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New post 03 Jul 2019, 21:03
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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
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Re: If xy > 0, is y/x + x/y > 2?   [#permalink] 03 Jul 2019, 21:03
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