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26 Jul 2018, 22:13
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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:
35%
(medium)
Question Stats:
72% (01:36) correct
28%
(01:25)
wrong
based on 181
sessions
History
Date
Time
Result
Not Attempted Yet
26 Jul 2018, 22:48
Kudos
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To determine if y is negative
x and y are real numbers
Statement 1
\(x^y\) > 0
x can be 10 and y can be 1 satisfying \(10^1 = 10 > 0\) and y is positive
x can be 10 and y can be -1 satisfying \(10^{-1} = \frac{1}{10} = 0.1 > 0\) and y is negative
Statement 1 is insufficient
Statement 2
\(y^x > 0\)
x can be 1 and y can be 10 satisfying \(10^1 = 10 > 0\) and y is positive
x can be 2 and y can be -10 satisfying \(-10^2 = 100 > 0\) and y is negative
Statement 2 is insufficient
Combining statements 1 and 2
x can be 10 and y can be 1 satisfying \(10^1 = 10 > 0\) and \(1^{10} = 1 > 0\) and y is positive
x can be 2 and y can be -1 satisfying \(2^{-1} = 0.5 > 0\) and \(-1^2 = 1 > 0\) and y is negative
Statements 1 and 2 together are insufficient
Hence option E
x and y are real numbers
Statement 1
\(x^y\) > 0
x can be 10 and y can be 1 satisfying \(10^1 = 10 > 0\) and y is positive
x can be 10 and y can be -1 satisfying \(10^{-1} = \frac{1}{10} = 0.1 > 0\) and y is negative
Statement 1 is insufficient
Statement 2
\(y^x > 0\)
x can be 1 and y can be 10 satisfying \(10^1 = 10 > 0\) and y is positive
x can be 2 and y can be -10 satisfying \(-10^2 = 100 > 0\) and y is negative
Statement 2 is insufficient
Combining statements 1 and 2
x can be 10 and y can be 1 satisfying \(10^1 = 10 > 0\) and \(1^{10} = 1 > 0\) and y is positive
x can be 2 and y can be -1 satisfying \(2^{-1} = 0.5 > 0\) and \(-1^2 = 1 > 0\) and y is negative
Statements 1 and 2 together are insufficient
Hence option E
26 Jul 2018, 23:16
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x and y are real numbers. Is y negative?
(1) \(x^y > 0\)
this can happen irrespective of value of y..
y is -1 and x is 2....x^y=2^(-1) = 1/2>0
y is 1 and x is 2....2^1=2>0
insuff
(2) y^x > 0
this can also happen with any value of y..
(-1)^2=1>0
1^2=1>0
insuff
combined..
y = -1 will satisfy both statements
y= 1 will satisfy both equations
insuff
E
(1) \(x^y > 0\)
this can happen irrespective of value of y..
y is -1 and x is 2....x^y=2^(-1) = 1/2>0
y is 1 and x is 2....2^1=2>0
insuff
(2) y^x > 0
this can also happen with any value of y..
(-1)^2=1>0
1^2=1>0
insuff
combined..
y = -1 will satisfy both statements
y= 1 will satisfy both equations
insuff
E
