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

Question

Competition Mode Question

Is 1/(y − x) > x − y ?

(1) y > x 
(2) |x − y| > 0

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AdiBir · Answer

Given: Nothing

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) .

ArunSharma12 · Answer

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 
e y , but we don't know whether x is greater than y or vice-versa?
not sufficient

Ans: A

HoneyLemon · Answer

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

bidskamikaze · Answer

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.

minustark · Answer

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.

kush09 · Answer

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
  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.

Preethi7777 · Answer

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)

mohitpandit · Answer

How do we know that the numerator will always have a positive value?

Preethi7777 · Answer

If it's negative it will be less than 0

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ArunSharma12 · Answer

1 + (y - x)^2 : minimum value for a square of any real number is zero.
1 + 0 : always positive

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

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