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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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03 Jul 2019, 08:00
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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, 08:10
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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

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

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03 Jul 2019, 08:14
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, 08:15
xy > 0 which means x and y should be of same sign

1) y= x- 1

Substituting in the x/y + y/x we get (2x^2-2x+1) / x(x-1) which can be further reduced to (2x-1)(x-1)/x(x-1) = (2x - 1) / x

So is (2x-1) / x > 2?

(2x-1) / x = 2 - (1/x) For all x >0 the equation will always be < 2
And for all x <0 the equation will always be > 2.

Hence 1 is Not Sufficient.

2) x = 2y

Substituting in the equation we get y / 2y + 2y / y = 2.5 [note: x and y are of same sign always as given in prompt ] . Hence the equation is >2

So statement 2 alone is sufficient. Hence option B.
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Re: If xy > 0, is y/x + x/y > 2?  [#permalink]

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03 Jul 2019, 08:18
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, 08:18
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, 08:20
1
If xy>0,
is y/x+x/y>2
(1) y=x−1
(2) x=2y

y/x + x/y = (x^2 + y^2)/xy
=> y/x+ x/y -2 >0
=> (x^2 + y^2 - 2xy)/xy>0
=> (x-y)^2/xy>0
=> xy>0 since (x-y)^2 is always >0 if x<>y and it is given that xy>0
The expression x/y+y/x>2 except when x=y

1. y = x- 1 => x<>y
(x-y)^2 = 1 > 0
Sufficient.

2. y = 2x
(x-y)^2 = x^2 > 0 since x<>0
Sufficient.

Each statement alone is sufficient.

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

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Updated on: 03 Jul 2019, 08:38
1
D, each independently sufficient.

x=2y
So, y/x + x/y = {y^2 + x^2}/xy
Substituting x=2y
{y^2 + 4*y^2}/2y^2, cancelling y^2 from numerator and denominator
(1+4)/2 > 2 sufficient

Statement#1
x=y+1, for y=1, x=2
2 + .5 is > 2
Also, for y=.1, x=1.1
11 + 1/11 is always >2
But
for y=-7, x=-6
7/6 + 6/7 >2,sufficient.
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Originally posted by LeoN88 on 03 Jul 2019, 08:20.
Last edited by LeoN88 on 03 Jul 2019, 08:38, edited 2 times in total.
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Re: If xy > 0, is y/x + x/y > 2?  [#permalink]

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03 Jul 2019, 08:23
Question stem =

[(x^2+y^2)/ xy] > 2

Using 2- we can easily get 2.5
by substituting y = x-1

using 1- we get
(y^2/y^2+1) +1 that may or may not be >2
hence not sufficient
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Re: If xy > 0, is y/x + x/y > 2?  [#permalink]

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03 Jul 2019, 08:24
1
Ans- D

1)
Substitute the value of y in the eq(n)

(x-1/x) + (x/x-1)
x cannot be 0 and 1 since denominator cannot be zero

Put x=2, value= 2.5
Put x=-1 or -2 value of eq(n) still >2
So sufficient

2) Substitute the value of x in terms of y

(y/2y)+(2y/y)
gives 2+1/2 which is greater than 2. So sufficient

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

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03 Jul 2019, 08:28
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]

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03 Jul 2019, 08:29
1
IMO D

If xy > 0, is y/x + x/y > 2?
Let's reorder the question:
y/x + x/y > 2 => x^2 + y^2 > 2xy (as xy > 0, we can easily multiply both sides by xy) => (x-y)^2 > 0 => x-y > 0 => x>y
Thus the question asks is x>y ?

St1: y=x-1 => if x=2, y=1 (note x or y cannot be 0 as given xy>0)
That means x>y, sufficient

St2: x=2y => if y=1, x=2
That means x>y, sufficient

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

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03 Jul 2019, 08:32
Statements 1 and 2 together are sufficient to answer the question.

Hence option C would be the answer.

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

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03 Jul 2019, 08:33
1
Either both x and y are negative or positive.

Lets start with second statement. Direct substitution will give us 2.5> 2

Statement 1 is a bit tricky. First simplify what the question is asking. y/x+x/y>2
LHS take common denominator and take it to RHS y^2 + x^2 > 2xy
This is basically simplified to (x - y)^2 > 0

Now replace statement 1 .... x - (x - 1) > 0 which is 1 > 0

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

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03 Jul 2019, 08:34
1
Given
xy>0 implies that x and y have same sign. Either both are +ve or both are -ve.

y/x+x/y>2 It is possible when x and y have sign, and when x and y are distinct real numbers.
We already know that x and y have same sign. If we know that x and y are distinct, we can give our answer.

Statement 1- x=y-1
x and y are distinct, hence statement is sufficient.

Statement 2- x=2y
Again we can see x and y are distinct, statement is sufficient

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

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03 Jul 2019, 08:34
Required is (x^2+y^2)/xy>0 when xy>0
(1) y=x−1 ; Insufficient as when substituted we get -2+1/x

(2) x=2y sufficient; 2+1/2 is greater than 2

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

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03 Jul 2019, 08:41
From the xy>0 >> x>0; y>0 or x<0; y<0

1) y=x-1
(x-1)/x+x/(x-1)>2
Not sufficient to answer the question

2) x=2y
y/2y+2y/y>2
1/2+2>2 Sufficient

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

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03 Jul 2019, 08:41
1
If xy>0, is y/x+x/y>2?

(1) y=x−1 ==>x-y = 1 ; (x-y)^2 = 1 ==> x^2 -2xy + y^2 = 1 > 0 ==> x^2+y^2 > 2xy => (x^2 + y^2)/xy > 2 (because xy > 0) ==> y/x + x/y > 2 --> yes
(2) x=2y ==> x/y = 2 & y/x = 1/2, so y/x+x/y = 2.5 > 2 --> yes

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

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03 Jul 2019, 08:41
Hi,
Given xy>0 means either x> 0, y>0 or x<0, y<0.
Now, the question:
$$\frac{(y)}{(x)}$$ + $$\frac{(x)}{(y)}$$ >2

=> ($$x^2$$ + $$y^2$$ -2*x*y) >0
=> $$(x-y)^2$$ >0
=> (x-y) >0
=> x > y ?

1. y = x-1
=> x = y+1
if x> 0 , y>0 then x > y.
if x<0 , y<0 then also x>y.
Sufficient.

2. x =2y
if x>0, y>0 then x>y.
if x<0, y<0 then x<y.
Insufficient.

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

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03 Jul 2019, 08:44
1
the given says that x,y ≠ zero, and have the same sign

the question can be rephrased to:
is $$\frac{y}{x} + \frac{x}{y} > 2$$
$$\frac{{y^2 + x^2}}{{xy}} > 2$$
$$y^2 + x^2 > 2xy$$
$$x^2 - 2xy + y^2 > 0$$
$$(x-y)^2 > 0?$$
is $$x≠y$$??

statement 1 says that (x-y) = 1, so yes $$(x-y)^2 > 0$$ --> sufficient
statement 2 says that x-y = y, so $$(x-y)^2 > 0$$ --> $$(y)^2 > 0$$ which is correct as y ≠ zero --> sufficient
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Re: If xy > 0, is y/x + x/y > 2?   [#permalink] 03 Jul 2019, 08:44

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