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09 Aug 2019, 02:37
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A
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:
65% (00:59) correct
35%
(01:08)
wrong
based on 58
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09 Aug 2019, 02:53
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Value of any negative square is always bigger hence from statement 1 we know that x and y are negative so A is sufficient.
Statement two x is greater than 1 and y is lower than 1 so if we take 2 as x and -3 as y the statement is satisfied but the sqaure of both number would flip inequities. Hence statement 2 is not sufficient
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Statement two x is greater than 1 and y is lower than 1 so if we take 2 as x and -3 as y the statement is satisfied but the sqaure of both number would flip inequities. Hence statement 2 is not sufficient
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09 Aug 2019, 02:59
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Is \(x^2\)>\(y^2\)?
STATEMENT 1--x > y > 0
From this we know x and y----positive and x>y
so ,\(x^2\)>\(y^2\)
when x = 2 y = 1
\(x^2\)>\(y^2\)
when x = \(\frac{1}{2}\) y = \(\frac{1}{3}\)
\(x^2\)>\(y^2\)
so , SUFFICIENT
STATEMENT 2--x > 1 > y
X>1 and y<1
when x =2 y = 0
then \(x^2\)>\(y^2\) --YES
when x = 2 y = -4
then \(x^2\)<\(y^2\) --NO
so INSUFFICIENT
A is the answer
STATEMENT 1--x > y > 0
From this we know x and y----positive and x>y
so ,\(x^2\)>\(y^2\)
when x = 2 y = 1
\(x^2\)>\(y^2\)
when x = \(\frac{1}{2}\) y = \(\frac{1}{3}\)
\(x^2\)>\(y^2\)
so , SUFFICIENT
STATEMENT 2--x > 1 > y
X>1 and y<1
when x =2 y = 0
then \(x^2\)>\(y^2\) --YES
when x = 2 y = -4
then \(x^2\)<\(y^2\) --NO
so INSUFFICIENT
A is the answer
