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

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Math Expert
Joined: 02 Sep 2009
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If xy > 0, is y/x + x/y > 2?  [#permalink]

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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) $$x = 2y$$

 This question was provided by Math Revolution for the Game of Timers Competition

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

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03 Jul 2019, 07:10
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$$\frac{x^2+y^2}{xy}>2$$

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

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

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03 Jul 2019, 21:03
3
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

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

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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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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

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

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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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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

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Re: If xy > 0, is y/x + x/y > 2?   [#permalink] 03 Jul 2019, 07:28