Is |x2+y2| > |x2-y2| ?
(1) x > y
(2) x > 0
Abs Value 2
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akhilsuhag wrote:Is |x2+y2| > |x2-y2| ?
(1) x > y
(2) x > 0
The answer to the question will almost always be yes, but if y can be 0, the answer can be no. So if you let, say, x=2 and y=1, then both statements are true, and the answer to the question is 'yes'. But if you let x=2 and y=0, then both statements are true, and the answer to the question is 'no'. So E is the answer.
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Alternate approach:akhilsuhag wrote:Is |x²+y²| > |x²-y²| ?
(1) x > y
(2) x > 0
When both sides of an inequality are enclosed in absolute value symbols, we can SQUARE the inequality.
|x²+y²|² > |x²-y²|²
x� + y� + 2x²y² > x� + y� - 2x²y²
4x²y² > 0
(xy)² > 0.
In the resulting inequality, the left side will always be positive as long as xy≠0.
Question stem, rephrased:
Is xy ≠0?
Statement 1: x > y
Case 1: x=1 and y=0
In this case, xy=0, and the answer to the rephrased question stem is NO.
Case 2: x=2 and y=1
In this case, xy≠0, and the answer to the rephrased question stem is YES.
Since the answer is NO in Case 1 but YES in Case 2, INSUFFICIENT.
Cases 1 and 2 satisfy BOTH STATEMENTS.
Since the answer is NO in Case 1 but YES in Case 2, the two statements combined are INSUFFICIENT.
The correct answer is E.
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A key takeaway from this problem is REMEMBERING that unless it's ruled out by information in either the question or the statements, the value of one or more variables in a DS question can be ZERO.
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Please suggest with what possible set of values we can evaluate such question
x,y ==> 2 1/2 0 -1/2 2
x,y ==> 2 1/2 0 -1/2 2
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Can't we write it as below considering both sides positive.
x2+ y2 > x2 - y2
2y2 >0
y2>0
which means y > 0 always so we can't take y=0 & hence explanation from GMATGuruNY is doubtful.
x2+ y2 > x2 - y2
2y2 >0
y2>0
which means y > 0 always so we can't take y=0 & hence explanation from GMATGuruNY is doubtful.
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Can't we write it as below considering both sides positive.
x2+ y2 > x2 - y2
2y2 >0
y2>0
which means y > 0 always so we can't take y=0 & hence explanation from GMATGuruNY is doubtful.
x2+ y2 > x2 - y2
2y2 >0
y2>0
which means y > 0 always so we can't take y=0 & hence explanation from GMATGuruNY is doubtful.