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Is |x – y| > |x| – |y|? (1) y < x (2) x y < 0

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Is |x – y| > |x| – |y|? (1) y < x (2) x y < 0  [#permalink]

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New post 26 Jan 2020, 08:53
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Question Stats:

82% (01:46) correct 18% (02:23) wrong based on 11 sessions

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Is \(|x – y| > |x| – |y|\)?
(1) \(y < x \)
(2) \(x y < 0\)
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Re: Is |x – y| > |x| – |y|? (1) y < x (2) x y < 0  [#permalink]

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New post 26 Jan 2020, 09:24
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naju wrote:
Is \(|x – y| > |x| – |y|\)?
(1) \(y < x \)
(2) \(x y < 0\)


Asked: Is \(|x – y| > |x| – |y|\)?

(1) \(y < x \)
If x=3;y=1; |x-y| =2; |x|-|y| = 2
But if x=3;y=-1; |x-y| =4; |x|-|y| = 2
Not sufficient

(2) \(x y < 0\)
x and y have opposite signs
If x=3; y=-2; |x-y| =5; |x|-|y| = 1
If x= -3; y=2; |x-y| = 5; |x|-|y|=1
|x-y| >|x|-|y|
Sufficient

IMO B

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Is |x – y| > |x| – |y|? (1) y < x (2) x y < 0  [#permalink]

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New post 26 Jan 2020, 09:37
naju wrote:
Is \(|x – y| > |x| – |y|\)?
(1) \(y < x \)
(2) \(x y < 0\)



Plugging in numbers is good way to approach this question. However, you may try and remember 2 properties that are quite useful:

#1. \(|x| + |y| ≥ |x + y|\)
Note: The '>' holds when x and y are of opposite signs i.e. one is positive and the other is negative
The '=' holds when x and y are of same signs i.e. both are positive or both are negative
Example: |5| + |2| = |5 + 2|; |-5| + |-2| = |-5 - 2|; |5| + |-2| > |5 - 2|


#2. \(|x| - |y| ≤ |x - y|\) (assuming |x| is greater than |y|)
Note: The '<' holds when x and y are of opposite signs i.e. one is positive and the other is negative
The '=' holds when x and y are of same signs i.e. both are positive or both are negative
Example: |5| - |2| = |5 - 2|; |-5| - |-2| = |-5 - (-2)|; |5| - |-2| < |5 - (-2)|


Using #2 above, it can be easily seen that statement 2 is sufficient:

Since xy < 0, it must be that x and y are of opposite signs

=> |x| - |y| < |x - y|

Answer B
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Re: Is |x – y| > |x| – |y|? (1) y < x (2) x y < 0  [#permalink]

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New post 26 Jan 2020, 19:15
Hi,
I followed the following approach to solve this question. Please correct me if I am wrong :-

Since it is DS, we need to simplify the question further if it is possible.
Is |x-y|>|x|-|y|?

Squaring both sides
Is \((x-y)^2\) > \((|x|-|y|)^2\)?
On further simplifying

Is \(x^2 -2xy+y^2\) > \(x^2-2|x||y|+y2\)?
=> Is xy>|x||y|?
=> Is xy>0?

So now we have to find out whether xy is greater than 0 or not.

Statement 1 => y<x
On re-arranging,
=> x-y>0
Now, we can take some values for x and y for which x-y is greater than 0,

let x = 2 , y= -4 => x - y = 6 > 0
\(xy = 2*(-4) = -8\) => xy <0

let x = 4, y= 2 => x - y = 2 > 0
\(xy = 4*(2) = 8\) => xy >0

We are not getting a definite answer from this statement. So this statement is insufficient.
We can eliminate options A and D.

Statement 2 => xy < 0
Now our simplified version of the question(Bolded Part) is that we have to find out : is xy > 0?
From this option we get that xy < 0.
We get a definite answer for the question : is xy > 0?
So this statement alone is sufficient.

Option B is the answer.
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Re: Is |x – y| > |x| – |y|? (1) y < x (2) x y < 0  [#permalink]

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New post 26 Jan 2020, 22:46
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Re: Is |x – y| > |x| – |y|? (1) y < x (2) x y < 0   [#permalink] 26 Jan 2020, 22:46
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