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

Question

If  xy > 0 , is  \frac{y}{x} + \frac{x}{y} > 2 ?

(1)  y = x - 1

(2)  x = 2y

firas92 · Accepted Answer

\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)

CareerGeek · Answer

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

Pls Hit Kudos if you like the solution

AvidDreamer09 · Answer

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

BelalHossain046 · Answer

Given, xy > 0 means neither x nor y can be ZERO

GMATinsight · Answer

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

Sayon · Answer

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   ->    
 
 (1) y = x -1 
=> x = y + 1
Substituting the above value in     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     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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If xy > 0, is y/x + x/y > 2?

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