|x + y| > |x| + |y|?
1) x + y < 0
2) xy < 0
Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient
Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
EACH statement ALONE is sufficient
Statements (1) and (2) TOGETHER are NOT sufficient
I think the answer is B.
If I take x = 0 and y = 1 i get 1 > 1 which is false.
However in statement 2 x and y cannot be 0 therefore one is positive and one is negative. making
x = 1 and y = -2
I get 1 < 3 sufficient. That is true for all vaules as one is always positive and the other negative and one never being zero.
Am I correct?
Abs Vaule Question
This topic has expert replies
From the question stem we can conclude that |x| + |y| is always +ve.
1) x+y<0
X = -2 and Y = -3 => X + Y = -5 <0; |X+Y| = 5; and |x| = 2 and |Y| = 3 ==> |X|+|Y| = 5
|X+Y|=|X|+|Y|
X = 2 and Y = -3 => X + Y = -1 <0; |X+Y| = 1; and |X| = 2 and |Y| = 3 ==> |X|+|Y| = 5
|X+Y|<|X|+|Y|
X = -3 and Y = 0 => X + Y = -3 <0; |X+Y| = 3 and |X| = 3 and |Y| = 0 |X| + |Y| = 3
|X+Y| = |X| + |Y|
So with statment one alone we are not getting fixed answer either YES or NO, hence it is not sufficient.
2) XY<0
hence either X or Y is -ve. Please note that both -ve and both +ve condition not possible as its prodcut results into +ve which inturn falls the given statement. Also ) is not possible, as the product XY results into 0 and inturn contradicts the given statement.
X = 2 and Y = -3 => X + Y = -1 => |X+Y| = 1 ; XY = -6 < 0 and |X| = 2 and |Y| = 3 ==> |X|+|Y| = 5 |X+Y|<|X|+|Y| So this concludes that |X+Y| is not greater than |X| + |Y|
X = -2 and Y = 3 => X + Y = 1 => |X+Y| = 1 ; XY = -6 < 0 and |X| = 2 and |Y| = 3 ==> |X|+|Y| = 5
|X+Y|<|X|+|Y| So this concludes that |X+Y| is not greater than |X| + |Y|
X = 5 and Y = -3 => X + Y = 2 => |X+Y| = 2 ; XY = -15 < 0 and |X| = 5 and |Y| = 3 ==> |X|+|Y| = 8
|X+Y|<|X|+|Y| So this concludes that |X+Y| is not greater than |X| + |Y|
Hence from above three examples it is clear that for any set of values of X & Y when either of the value is -ve |X+Y| is always less than |X| + |Y|. This is true for fractions also.
Hence for all combinations we get only one result i.e. NO which is sufficient.
Hence statement 2 alone is sufficient to answer this question. Hence the answer is B.
1) x+y<0
X = -2 and Y = -3 => X + Y = -5 <0; |X+Y| = 5; and |x| = 2 and |Y| = 3 ==> |X|+|Y| = 5
|X+Y|=|X|+|Y|
X = 2 and Y = -3 => X + Y = -1 <0; |X+Y| = 1; and |X| = 2 and |Y| = 3 ==> |X|+|Y| = 5
|X+Y|<|X|+|Y|
X = -3 and Y = 0 => X + Y = -3 <0; |X+Y| = 3 and |X| = 3 and |Y| = 0 |X| + |Y| = 3
|X+Y| = |X| + |Y|
So with statment one alone we are not getting fixed answer either YES or NO, hence it is not sufficient.
2) XY<0
hence either X or Y is -ve. Please note that both -ve and both +ve condition not possible as its prodcut results into +ve which inturn falls the given statement. Also ) is not possible, as the product XY results into 0 and inturn contradicts the given statement.
X = 2 and Y = -3 => X + Y = -1 => |X+Y| = 1 ; XY = -6 < 0 and |X| = 2 and |Y| = 3 ==> |X|+|Y| = 5 |X+Y|<|X|+|Y| So this concludes that |X+Y| is not greater than |X| + |Y|
X = -2 and Y = 3 => X + Y = 1 => |X+Y| = 1 ; XY = -6 < 0 and |X| = 2 and |Y| = 3 ==> |X|+|Y| = 5
|X+Y|<|X|+|Y| So this concludes that |X+Y| is not greater than |X| + |Y|
X = 5 and Y = -3 => X + Y = 2 => |X+Y| = 2 ; XY = -15 < 0 and |X| = 5 and |Y| = 3 ==> |X|+|Y| = 8
|X+Y|<|X|+|Y| So this concludes that |X+Y| is not greater than |X| + |Y|
Hence from above three examples it is clear that for any set of values of X & Y when either of the value is -ve |X+Y| is always less than |X| + |Y|. This is true for fractions also.
Hence for all combinations we get only one result i.e. NO which is sufficient.
Hence statement 2 alone is sufficient to answer this question. Hence the answer is B.
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I think @seal4913 you must have got the question wrong.
Because,
always, |x+y| <= |x| + |y| (the 2 sides are equal when both x and y have the same sign and RHS<LHS when x and y have opposite signs)
Hence Is |x+y| > |x| + |y|? is always NO
So in that case your answer should be D
But if your question is
Is |x+y| < |x| + |y|? then B is the answer as explained by pootej
Because,
always, |x+y| <= |x| + |y| (the 2 sides are equal when both x and y have the same sign and RHS<LHS when x and y have opposite signs)
Hence Is |x+y| > |x| + |y|? is always NO
So in that case your answer should be D
But if your question is
Is |x+y| < |x| + |y|? then B is the answer as explained by pootej
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Hi, there. I'll add my two cents to this discussion.
I completely agree with Mr Smith. The original question, as posted by seal4913, is a faulty DS questions, because the prompt question always has an answer of no, irrespective of the two statements. No legitimate GMAT DS question will have a prompt that is always true or always false and therefore entirely independent of the two statements.
Just to review the math facts
|x| + |y| = |x + y| only when x and y have the same sign --- both positive or both negative (or, either one of them could be zero)
|x| + |y| > |x + y| when x and y have opposite sign.
In the original DS question, if the prompt were |x + y| < |x| + |y|, instead of |x + y| > |x| + |y|, then it would have been a sensible DS question with an answer of B, as Mr Smith has said.
Just for further practice, here's a PS problem with absolute value:
https://gmat.magoosh.com/questions/131
When you submit your answer to that question, the next page will have a video explanation. Each one of 800+ of Magoosh's GMAT questions has its own video explanation.
I hope all that helps. Let me know if anyone has any questions on what I've written.
Mike
I completely agree with Mr Smith. The original question, as posted by seal4913, is a faulty DS questions, because the prompt question always has an answer of no, irrespective of the two statements. No legitimate GMAT DS question will have a prompt that is always true or always false and therefore entirely independent of the two statements.
Just to review the math facts
|x| + |y| = |x + y| only when x and y have the same sign --- both positive or both negative (or, either one of them could be zero)
|x| + |y| > |x + y| when x and y have opposite sign.
In the original DS question, if the prompt were |x + y| < |x| + |y|, instead of |x + y| > |x| + |y|, then it would have been a sensible DS question with an answer of B, as Mr Smith has said.
Just for further practice, here's a PS problem with absolute value:
https://gmat.magoosh.com/questions/131
When you submit your answer to that question, the next page will have a video explanation. Each one of 800+ of Magoosh's GMAT questions has its own video explanation.
I hope all that helps. Let me know if anyone has any questions on what I've written.
Mike
Magoosh GMAT Instructor
https://gmat.magoosh.com/
https://gmat.magoosh.com/