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Is 1/(y − x) > x − y ? (1) y > x (2) |x − y| > 0

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Bunuel
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Is 1/(y − x) > x − y ? (1) y > x (2) |x − y| > 0 [#permalink] New post   10 Aug 2020, 00:51
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A
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D
E

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55% (hard)

Question Stats:

62% (01:48) correct 38% (02:02) wrong based on 152 sessions
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Is 1/(y − x) > x − y ?

(1) y > x
(2) |x − y| > 0


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A
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B
AdiBir
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Re: Is 1/(y − x) > x − y ? (1) y > x (2) |x − y| > 0 [#permalink] New post   10 Aug 2020, 04:02
2
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Given:
Nothing :lol:
To Find:
Is \(\frac{1}{(y-x)} > x-y\)

Pre Process:
\(\frac{1}{(y-x)} > x-y\)
\(\frac{1}{(y-x)} - (x-y) >0\)
\(\frac{1}{(y-x)} + (y-x) >0\)
\(\frac{1 + (y-x)^2}{(y-x)(} >0\)

Since numerator is always positive, we need to find whether denominator is positive.
\(y-x > 0\)
y >x ?

Now coming to the statements:

1) \(y > x.\)
Just what we need. So given expression is positive.
1) is Sufficient.

2) \(|x-y| > 0.\)
The only thing this tells us is that x is not equal to y.
But no information about \(y>x.\)
2) is not Sufficient.

Hence correct option is A).
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Re: Is 1/(y − x) > x − y ? (1) y > x (2) |x − y| > 0 [#permalink] New post   10 Aug 2020, 04:25
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imo A
Is 1/(y − x) > x − y ?

(1) y > x
so y-x >0 .

so 1/(y − x) > x − y ...yes ..
(positive ) > -ve -- Yes .. Sufficient

(2) |x − y| > 0
no information whether x>y or y>x .
so not sufficient .

So ans is A
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Re: Is 1/(y − x) > x − y ? (1) y > x (2) |x − y| > 0 [#permalink] New post   10 Aug 2020, 15:25
1
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Is 1/(y − x) > x − y ?

(1) y > x
(2) |x − y| > 0

1) y = .2, x = .1, then 1/(.2-.1) = 10 and x - y = -0.1. ok. Now, when y = -5, x = -6, then LHS = 1, RHS = -1. ok. Now, lets take different sign, y =5, x = -2, LHS = 1/7, RHS = -7. So, 1/ (y-x) is always greater than x - y. sufficient.

2) if x = -2, y = -3, then LHS = -1, RHS = 1. Again from the explanations of statement 1, we can say LHS > RHS. No unique possibilities. Not sufficient.

A is the answer.
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Re: Is 1/(y − x) > x − y ? (1) y > x (2) |x − y| > 0 [#permalink] New post   10 Aug 2020, 20:58
1
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Is 1/(y − x) > x − y ?

Solution:
=> 1/(y − x) > x − y
=> 1/(y − x) (x − y) > 1
=> 1 /(2xy-x^2-y^2)>1
=> 1/-(x+y)^2 >1
=> -1/(x+y)^2>1


For any value of x,y: (x+y)^2 will always be positive and -1/(x+y)^2>1 will be false; only exception is x=y=0.

(1) y > x
Considering statement 1 alone.
x and y cannot be equal, that means both cannot be 0, hence unique answer is possible.

(2) |x − y| > 0
[color=#0000ff]Considering statement 2 alone.
x =4, y =2 ; we will get -1/(x+y)^2>1 as NO.
however, for x=2 and y = -2; -1/(x+y)^2>1 will be undefined.
Statement 2 alone is not possible.

IMO A

PS: while solving I selected answer as D, but during writing solution I realized 2 condition cannot give definite answer.
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Re: Is 1/(y − x) > x − y ? (1) y > x (2) |x − y| > 0 [#permalink] New post   10 Aug 2020, 22:15
1
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given in the stem :
1/y-x > x-y
1/y-x > -(y-x)
1/y-x + y-x > 0
1 + (y-x)2 / (y-x) > 0...........(1)
we know that numerator is positive as (y-x)2 is non negative and (y-x)2 + 1 will thus be positive
hence, for (1) to be true , given that numerator is positive , denominator must be positive
therefore, question can be simplified as : is (y-x) > 0

now statement 1 :
y>x
y-x>0
SUFFICIENT

statement 2 :
| x-y | > 0
ex 1 : x=4 y=2 | 4-2 | > 0
ex 2 : x=-4 y=2 |-4-2| > 0
hence x>y or y>x , both satisfy the equation
INSUFFICIENT

ANSWER (A)
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Re: Is 1/(y − x) > x − y ? (1) y > x (2) |x − y| > 0 [#permalink] New post   10 Aug 2020, 04:05
Quote:
Is 1/(y − x) > x − y ?

(1) y > x
(2) |x − y| > 0


Given:
\(\frac{1}{(y − x)} > x − y \)
\(\frac{1}{(y − x)} + y - x > 0 \)
\(\frac{1 + (y − x)(y - x)}{y - x} > 0 \)
\(\frac{1 + (y − x)^2 }{ y -x} > 0 \)

statement 1: y > x
(y - x)^2 is a non-negative value.

numerator will always be a positive value.
as y > x, denominator will be positive irrespective of the sign of y and x.

sufficient

statement 2: |x - y| > 0
\(x \ne y\), but we don't know whether x is greater than y or vice-versa?
not sufficient

Ans: A
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Re: Is 1/(y − x) > x − y ? (1) y > x (2) |x − y| > 0 [#permalink] New post   10 Aug 2020, 10:34
1) y > x
consider x = -2, y = -1, substitute in inequality we get
1 > -1 (Yes, greater)
consider x = 1, y = 2, substitute in inequality we get
1 > 1 (No, not greater)
Yes and No both are possible, insufficient.

2) \(\mid{x-y}\mid\) > 0
This gives us:
For x-y < 0,
y-x<0, so y<x;
and for \(x-y\geq{0}\) also,
y<x.
Now consider x= -1, y= -2, substitute in inequality we get
-1 > 1 (No, not greater)
consider x=2, y=1, substitute in inequality we get
-1 > 1 (No, not greater)
B gives NO as answer in both cases.

Answer B.
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Re: Is 1/(y − x) > x − y ? (1) y > x (2) |x − y| > 0 [#permalink] New post   13 Aug 2020, 03:56
How do we know that the numerator will always have a positive value?
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Re: Is 1/(y − x) > x − y ? (1) y > x (2) |x − y| > 0 [#permalink] New post   13 Aug 2020, 04:10
If it's negative it will be less than 0

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Re: Is 1/(y − x) > x − y ? (1) y > x (2) |x − y| > 0 [#permalink] New post   13 Aug 2020, 04:15
mohitpandit wrote:
How do we know that the numerator will always have a positive value?



\(1 + (y - x)^2\): minimum value for a square of any real number is zero.
1 + 0 : always positive
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Re: Is 1/(y − x) > x − y ? (1) y > x (2) |x − y| > 0 [#permalink]
13 Aug 2020, 04:15
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