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Difficulty:
55%
(hard)
Question Stats:
62% (01:48) correct
38%
(02:03)
wrong
based on 154
sessions
History
Date
Time
Result
Not Attempted Yet
Is 1/(y − x) > x − y ?
(1) y > x
(2) |x − y| > 0
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Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
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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).
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
Joined: 24 Oct 2015
Last visit: 20 Jul 2022
Posts: 512
Location: India
Schools: IIML-IPMX '22 (A)
GMAT 1: 650 Q48 V31
GMAT 2: 720 Q49 V38 (Online)
GPA: 4
Products:
Quote:
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
HoneyLemon
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Posts: 627
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
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
Joined: 07 Jan 2018
Last visit: 29 Oct 2022
Posts: 251
Location: India
Schools: ESMT NUS '22 IIMA PGPX'23 NTU
GMAT 1: 710 Q49 V38
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.
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
Joined: 14 Jul 2019
Last visit: 01 Apr 2021
Posts: 465
Status:Student
Location: United States
Concentration: Accounting, Finance
Schools: Zicklin Baruch "22
GMAT 1: 650 Q45 V35
GPA: 3.9
WE:Education (Accounting)
Products:
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.
(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.
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.
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.
Posts: 5
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)
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)
Posts: 1
How do we know that the numerator will always have a positive value?
Posts: 5
If it's negative it will be less than 0
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ArunSharma12
Joined: 24 Oct 2015
Last visit: 20 Jul 2022
Posts: 512
Location: India
Schools: IIML-IPMX '22 (A)
GMAT 1: 650 Q48 V31
GMAT 2: 720 Q49 V38 (Online)
GPA: 4
Products:
mohitpandit
\(1 + (y - x)^2\): minimum value for a square of any real number is zero.
1 + 0 : always positive
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