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27 Jun 2017, 00:39
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
41% (02:08) correct
59%
(01:48)
wrong
based on 224
sessions
History
Date
Time
Result
Not Attempted Yet
If x and y are integers, is x > y?
(1) \(y^x < −1\)
(2) \(\frac{x}{|y|} > 1\)
(1) \(y^x < −1\)
(2) \(\frac{x}{|y|} > 1\)
27 Jun 2017, 01:07
Kudos
Bookmarks
Bunuel
Statement 1: implies that y<0 and "x" is Odd Positive integer because if x is odd negative integer then in that case y^x = 1/y^x which will be greater than -1. For eg. if x = -3 & y = -2 then (-2)^(-3) = 1/(-2)^3 = -1/8 > -1. Hence "x" is odd positive integer so x>y. Sufficient
Statement 2: x/|y|>1, or x>|Y|. therefore "x" is positive and its magnitude is greater than "y". Hence sufficient.
Option D
27 Jun 2017, 01:10
Kudos
Bookmarks
Bunuel
Given : x and y are integers
DS: x > y
Option 1 : y^x < -1 , Since x and y are integers y<-1 and x must be odd and +ve. So x>y
Option 2: x/|y| >1
x>|y|
If y>0, x>y
If y<0 , x>-y i.e. x is +ve and y is -ve . So x>y.
Both options individually solves the problem
So Option D.
