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Is |x| + |y| > |x - y| ?

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Bunuel
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Is |x| + |y| > |x - y| ? [#permalink] New post  06 Jun 2016, 04:10
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Is \(|x| + |y| > |x - y|\) ?

(1) \(|x| > |y|\)

(2) \(|x - y| < |x|\)
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Re: Is |x| + |y| > |x - y| ? [#permalink] New post  06 Jun 2016, 06:26
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|x| + |y| > |x - y| when x and y are of same sign and not equal to 0.

St1: | x | > | y | --> We cannot infer whether both x and y are of same sign. Not Sufficient

St2: | x - y | < | x |
Square on both sides --> x^2 + y^2 - 2xy < x^2
y^2 - 2xy < 0
y(y - 2x) < 0
If y is positive --> y - 2x = -ve --> x is positive
If y is negative --> y - 2x = +ve --> x is negative
x and y are of same sign.
Sufficient

Answer: B
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Re: Is |x| + |y| > |x - y| ? [#permalink] New post  06 Jun 2016, 08:09
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I used the approach of trying to prove that there are both YES and NO answers to every statement.

From just observing the given equation I concluded that both x & y have to have the same sign and not equal 0.

1) | x | > | y |
x = 4, y = 3 (same sign)
| x | > | y |

x = -4, y = 3 (different sign)
| x | > | y |

NOT SUFFICIENT

2) | x - y | < | x | or | x | > | x - y |
x = 4, y = 3 (same sign)
| x | > | x - y |

x= -4. y = -3
| x | > | x - y |
Unsuccessfully yrying to think of a number that fits this equation when x & y have opposite signs, I concluded that this is sufficient to answer NO to the posed question.
SUFFICIENT

Answer: B
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Re: Is |x| + |y| > |x - y| ? [#permalink] New post  06 Jun 2016, 09:56
St1:- does not hold for y<0 & x>0 (x=1 & y=-1). But holds true for x=y=1 so insufficient.
St2:- always holds true as
|x-y|<|x| and addition of any positive number ( |y| ) will not affect the inequality.

Hence answer is 'B'

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Divyadisha
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Re: Is |x| + |y| > |x - y| ? [#permalink] New post  30 Jun 2016, 11:50
Bunuel wrote:
Is | x | + | y | > | x - y | ?

(1) | x | > | y |

(2) | x - y | < | x |


| x | + | y | > | x - y |

My understanding of the question is that if either x or y is -ve | x - y | < | x | + | y | (because x and y will both be +ve on RHS)

(1) | x | > | y |

In this case x or y can be -ve or +ve. Not sufficient.

(2) | x - y | < | x |

Since, | y | will always be +ve, adding it to | x | will yield a +ve number > | x - y |

Sufficient

Answer is B
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Is |x|+|y|>|x - y|? [#permalink] New post  06 Apr 2017, 19:39
Is \(│x│+│y│>│x - y│\) ?

(1) \(│x│>│y│\)
(2) \(│x-y│ < │x│\)

Please help to resolve. I have a hard time to solve this problem. Is it possible to solve by value testing for Statement (2)?
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ar500
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Re: Is |x|+|y|>|x - y|? [#permalink] New post  06 Apr 2017, 20:21
ziyuen wrote:
Is \(│x│+│y│>│x - y│\) ?

(1) \(│x│>│y│\)
(2) \(│x-y│ < │x│\)

Please help to resolve. I have a hard time to solve this problem. Is it possible to solve by value testing for Statement (2)?


Statement 1:
If x and y are positive then the question holds, however if x is negative and y is positive it does not. For example, (-5,2) yields an equality. Accordingly, statement 1 is insufficient.

Statement 2:
Solve using transitive property and property of absolute values. Given │x-y│ < │x|, we just need to test to see if |x|+|y| < |x|; as long as it isn't, statement 2 is sufficient. Since we're dealing with absolute values here there is no way that |x|+|y| < |x| because |y| can't be less than zero.
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Re: Is |x|+|y|>|x - y|? [#permalink] New post  06 Apr 2017, 21:00
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Is \(|x| + |y| > |x - y|\)?

Square on both sides, \(x^2 + y^2 + 2|x||y| > x^2 + y^2 - 2xy\)?
Question: \(|x||y| > -(xy)\)?
LHS is positive but RHS may be positive or negative.
For example lets take a case where |x| = 2 |y| = 3 but x = 2 and y = -3 --> In this case \(|x||y| = -(xy)\)
but if x and y are positive then \(|x||y| > -(xy)\)

St1: \(│x│>│y│\)
x = 3; y = -2 --> Is \(|x||y| > -(xy)\)? No
x = 3; y = 2 --> Is \(|x||y| > -(xy)\)? Yes
Not Sufficient

St2: \(│x−y│<│x│\)
Square on both sides, \(x^2 + y^2 - 2xy < x^2\)
\(y^2 < 2xy\)
\(xy > \frac{y^2}{2}\) --> xy is positive.
Hence \(|x||y| > -(xy)\)
Sufficient

Answer: B
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Re: Is |x| + |y| > |x - y| ? [#permalink] New post  06 Apr 2017, 22:53
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ziyuen wrote:
Is \(│x│+│y│>│x - y│\) ?

(1) \(│x│>│y│\)
(2) \(│x-y│ < │x│\)

Please help to resolve. I have a hard time to solve this problem. Is it possible to solve by value testing for Statement (2)?


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