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10 Sep 2017, 21:42
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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:
54% (01:38) correct
46%
(01:52)
wrong
based on 337
sessions
History
Date
Time
Result
Not Attempted Yet
10 Sep 2017, 23:14
Kudos
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Is y^x > x^y?
(1) y > x
if y=2 x=1 => 2^1 > 1^ 2 => 2>1 true
y=2 x= -2 => 2^(-2) > (-2)^2 => 1/4 > 2 False
Not sufficient
(2) x is a positive number
so x>0 . Nothing about y.
if y=2 x=1 => 2^1 > 1^ 2 => 2>1 true
if y =-2 x=1 => (-2)^1 > (1)^(-2) => -2 > 1 false
So Not sufficient
1+2
y> x and x>0
( Its not given that numbers are integers so always check for fractions. If given x and y are integers equation might be different)
if y=2 x=1 => 2^1 > 1^ 2 => 2>1 true
y= 1/2 and x=1/4 => (1/2)^(1/4) > (1/4)^(1/2) => (1/2)^(1/4) > 1/2 True ( Well this step not required)
y =5 and x=3 => 5^ 3 > 3^5 => 125 > 243 False
Not Sufficient
Answer: E
(1) y > x
if y=2 x=1 => 2^1 > 1^ 2 => 2>1 true
y=2 x= -2 => 2^(-2) > (-2)^2 => 1/4 > 2 False
Not sufficient
(2) x is a positive number
so x>0 . Nothing about y.
if y=2 x=1 => 2^1 > 1^ 2 => 2>1 true
if y =-2 x=1 => (-2)^1 > (1)^(-2) => -2 > 1 false
So Not sufficient
1+2
y> x and x>0
( Its not given that numbers are integers so always check for fractions. If given x and y are integers equation might be different)
if y=2 x=1 => 2^1 > 1^ 2 => 2>1 true
y= 1/2 and x=1/4 => (1/2)^(1/4) > (1/4)^(1/2) => (1/2)^(1/4) > 1/2 True ( Well this step not required)
y =5 and x=3 => 5^ 3 > 3^5 => 125 > 243 False
Not Sufficient
Answer: E
Mo2men
Joined: 26 Mar 2013
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11 Sep 2017, 02:21
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Bunuel
(1) y > x
Let y=2 & x =1.........2^1 > 1^2...........Answer is yes
Let y=4 & x =3.........4^3 > 3^4...........Answer is NO
Insufficient
(2) x is a positive number
Same examples above
Insufficient
Combine 1 & 2
Same examples above
Insufficient
Answer: E
