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07 Apr 2023, 09:45
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
51% (01:59) correct
49%
(02:00)
wrong
based on 53
sessions
History
Date
Time
Result
Not Attempted Yet
Is \(x^2-y^2 < x-y?\)
(1) \(x - y < 0\)
(2) \(x + y > 0\)
(1) \(x - y < 0\)
(2) \(x + y > 0\)
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
07 Apr 2023, 10:08
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Bunuel
\(x^2 - y^2 < x - y\)
\((x + y)(x - y)< x - y\)
\((x - y)(x + y - 1)< 0\)
The above inequality holds true under the below conditions -
- (x - y) < 0
(x+y) > 1 - (x - y) > 0
(x+y) < 1
Statement 1
(1) \(x - y < 0\)
We are presented with information of (x-y), we don't know whether (x+y) > 1. Hence, without that information, we cannot find out a definite answer to the question Is \(x^2-y^2 < x-y\)?
The information alone is not sufficient, and we can eliminate A and D.
Statement 2
(2) \(x + y > 0\)
We are presented with information of (x+y), we don't know whether (x-y) > 0 and whether (x + y) > 1. Hence, without that information, we cannot find out a definite answer to the question Is \(x^2-y^2 < x-y\)?
The information alone is not sufficient, and we can eliminate B.
Combined
The statements combined provide us with the below information -
(1) \(x - y < 0\)
(2) \(x + y > 0\)
The statements combined don't help either as we still don't know if (x + y) > 1.
Option E
08 Apr 2023, 09:45
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given
\(x^2-y^2 < x-y?\)
(x+y)(x-y)-(x-y)<0
target is (x-y)*[(x+y)-1] <0
#1
\(x - y < 0\)
x has to be less than y if y is +ve (-1,5)and
x is less than y it both are -ve (-2,-1)
also (0,1) we get no to target
insufficient
#2 \(x + y > 0\)
check with x,y : (1,2 ) we get yes to target and at x,y ( 0,1) we get no to target
from 1 &2
possible values of x,y
(0,1) and ( 1,2)
insufficient
option E
\(x^2-y^2 < x-y?\)
(x+y)(x-y)-(x-y)<0
target is (x-y)*[(x+y)-1] <0
#1
\(x - y < 0\)
x has to be less than y if y is +ve (-1,5)and
x is less than y it both are -ve (-2,-1)
also (0,1) we get no to target
insufficient
#2 \(x + y > 0\)
check with x,y : (1,2 ) we get yes to target and at x,y ( 0,1) we get no to target
from 1 &2
possible values of x,y
(0,1) and ( 1,2)
insufficient
option E
Bunuel
