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Statement II is sufficient xy<0 means that either one of the numbers is negative and other is positive

|x-y| will always result to be a greater value as negative - positive or positive - negative will result in addition of the numbers and |x| - |y| will always result in subtraction.

Bunuel, for 1.B when .. y ..0 .. x, you said \(|x-y|>|x|-|y|\) becomes: \(x-y>x+y\). and 1.c when ... y ... x ... 0, you said \(|x-y|>|x|-|y|\) becomes: \(x-y>-x+y\) --> \(x>y\).

Can you explain this a little bit more? How did you go about removing the absolute signs for this scenarios?

Thanks for your help.

Consider absolute value of some expression - \(|some \ expression|\): If the expression in absolute value sign (||) is negative or if \(some \ expression<0\) then \(|some \ expression|=-(some \ expression)\); If the expression in absolute value sign (||) is positive or if \(some \ expression>0\) then \(|some \ expression|=some \ expression\).

(It's the same as for \(|x|\): if \(x<0\), then \(|x|=-x\) and if \(x>0\), then \(|x|=x\))

We have \(|x-y|>|x|-|y|\):

For B: ---------\(y\)---\(0\)---------\(x\)---, \(y<0<x\) (\(x>y\)) --> so as \(x-y>0\), then \(|x-y|=x-y\). Also as \(x>0\), then \(|x|=x\) and as \(y<0\), then \(|y|=-y\). So in this case \(|x-y|>|x|-|y|\) becomes: \(x-y>x-(-y)\) or \(x-y>x+y\) --> \(2y<0\) --> \(y<0\). Which is right, as we consider the range \(y<0<x\);

The same for C.

Hope it's clear.

Hi Bunuel, Thanks for this great explanation. But I am still unclear about this.

you said... for the case B.. \(y<0<x\) (\(x>y\)) --> so as \(x-y>0\), then \(|x-y|=x-y\). I understand this.

but how is this the same when it comes to case C. where x and y both are negative (y<x<0), although i do get that X is still greater than Y but I am confused how would it still translate to \(|x-y|=x-y\) when both of them are negative... wouldnt it be more like \(|x-y|=y-x\) , i get the RHS part ... its the LHS where I am confused.

please explain thanks!

Consider x=-1 and y=-2 for C: in this case \(|x-y|=|-1-(-2)|=1=x-y\).

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

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24 Jan 2014, 13:17

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Squaring and simplifying, is xy < |x||y|? 1. x > y. put x = -2, y = -4 NO; put x = 4, y = -2 YES. NOT SUFFICIENT 2. xy <0; |x||y| always > 0, so xy always < |x||y| SUFFICIENT

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

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28 Jun 2014, 19:32

First of all we need to consider different cases to solve this problem. take option 1) y<x this option can be subdivided into two blocks...when both are x,y>0 and x>y. lets take x=2, y=1 lx-yl = l2 -1l = 1 Right hand side of the equation = lxl - lyl = l2l - l1l = 1....so equation is invalid. lets take another example when x= 1 and y = -1... lx-yl = l1 - (-1)l = 2 and Rgiht hand side = 0 which make our equation valid....hence we cannot conclude anything from this option.

take option 2) xy<0 under this option there can be two cases....a) x>0 and y<0 (b) x<0 and y>0 lets take a) and use some values.... x=2 and y = -1... simplifing the equation we get...lx-yl = 3 where lxl - lyl = 1 it makes equation valid. now take b) x= -2 and y = 1...we get lx-yl = 3 and lxl - lyl = 1 its also satisfy our given equation. so this option is sufficent to answer the given question.

(II) For all real x and y, \(|x - y| \geq |x| - |y|\) \(|x - y| = |x| - |y|\) when (1) x and y have the same sign and x has greater (or equal) absolute value than y (2) y is 0 \(|x - y| > |x| - |y|\) in all other cases

Question: Is |x-y|>|x|-|y|?

We need to establish whether the "equal to" sign can hold or not.

(1) y < x Doesn't tell us whether they have the same sign or opposite. So we don't know whether the equal to sign will hold or greater than. Not sufficient.

(2) xy < 0 Tells us that one of x and y is positive and the other is negative (they do not have same sign). Also tells us that neither x nor y is 0. Hence, the "equal to" sign cannot hold. Sufficient to answer 'Yes'

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

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11 Jan 2015, 05:48

I have recently started with the prep thus looking for patterns to solve the Absolute Value questions. Wanted to check if the below solution stands valid.

Square both the sides of the question stem |x-y|^2>(|x|-|y|)^2 --- |x-y|^2 = (x-y)^2 x^2+y^2-2xy>x^2+y^2-2|x||y| xy<|x||y|

From 1 y<x - In Sufficient

From 2 xy<0 (Either of them is negative i.e. x +ve y -ve or y +ve or x -ve) Sufficient

I have recently started with the prep thus looking for patterns to solve the Absolute Value questions. Wanted to check if the below solution stands valid.

Square both the sides of the question stem |x-y|^2>(|x|-|y|)^2 --- |x-y|^2 = (x-y)^2 x^2+y^2-2xy>x^2+y^2-2|x||y| xy<|x||y|

From 1 y<x - In Sufficient

From 2 xy<0 (Either of them is negative i.e. x +ve y -ve or y +ve or x -ve) Sufficient

Note that "Is a > b" is not the same question as "Is a^2 > b^2?"

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

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30 May 2015, 16:30

kancharana wrote:

mmphf wrote:

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

(1) y < x (2) xy < 0

How it is B? Did they mention that X and Y are integers? No right, the answer should be E. If they provide details about X and Y as integers then it will be B otherwise it will be E.

can anyone help me about the scenario whether we consider fractions or not in this case?

Scenario:

x=1/2, y=1/3 ==> |1/2-1/3|=1/6 and |1/2|-|1/3|=1/6

You're forgetting that in statement 2, given that xy<0, either x or y need to be negative.

Scenario 1: x is negative |-1/2 - 1/3| = |-5/6| = 5/6 ....other side of the equation....|-1/2|-|1/3| = 1/6 5/6 > 1/6

Scenario 2: y is negative |1/2-(-1/3)| = 5/6....other side of the equation....|1/2|-|-1/3| = 1/2-1/3 = 1/6 5/6 > 1/6

In both scenarios |x-y| is greater than |x|-|y|. This would also be true if you were to switch the values so x = 1/3 and y = 1/2. So statement 2 is sufficient

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

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11 Oct 2015, 07:17

mmphf wrote:

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

(1) y < x (2) xy < 0

A great way to rember the rules Bunuel mentions for me is to remember what happens when you have a difference between a single absolute value term |X-Y| and a difference between two seperate absolute values |x| - |y|. So in order for |x-y| to be greater than |x| - |y|, xy need to have different signs.

Why is this? Because if xy<0, xy add to each other in every case within this term |x - y| while within |x| - |y| the whole value of the RHS shrinks.
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Re: Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 [#permalink]

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14 Oct 2015, 18:47

Bunuel wrote:

someone79 wrote:

This is a very simple question. !x-y!>|x|-|y| can only happen if both the numbers are of different signs.

If xy<0 then these numbers are of opposite signs. Hope this clears.

X=2 y=3 then |x-y|=|x|-|y| if x=-2 and y = 3 then |x-y|>|x|-|y|

Red part is not correct \(|x-y|>{|x|-|y|}\) also holds true when \(x\) and \(y\) have the same sign and the magnitude of \(y\) is more than that of \(x\) (so for \(|y|>|x|\)). Example: \(x=2\) and \(y=3\) --> \(|x-y|=1>-1={|x|-|y|}\); \(x=-2\) and \(y=-3\) --> \(|x-y|=1>-1={|x|-|y|}\).

Actually the only case when \(|x-y|>{|x|-|y|}\) does not hold true is when \(xy>{0}\) (so when \(x\) and \(y\) have the same sign) and \(|x|>|y|\) (simultaneously). In this case \(|x-y|={|x|-|y|}\) (as shown in my previous post). Example: \(x=3\) and \(y=2\) --> \(|x-y|=1={|x|-|y|}\); \(x=-3\) and \(y=-2\) --> \(|x-y|=1={|x|-|y|}\).

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

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15 Oct 2015, 03:43

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longfellow wrote:

Bunuel wrote:

someone79 wrote:

This is a very simple question. !x-y!>|x|-|y| can only happen if both the numbers are of different signs.

If xy<0 then these numbers are of opposite signs. Hope this clears.

X=2 y=3 then |x-y|=|x|-|y| if x=-2 and y = 3 then |x-y|>|x|-|y|

Red part is not correct \(|x-y|>{|x|-|y|}\) also holds true when \(x\) and \(y\) have the same sign and the magnitude of \(y\) is more than that of \(x\) (so for \(|y|>|x|\)). Example: \(x=2\) and \(y=3\) --> \(|x-y|=1>-1={|x|-|y|}\); \(x=-2\) and \(y=-3\) --> \(|x-y|=1>-1={|x|-|y|}\).

Actually the only case when \(|x-y|>{|x|-|y|}\) does not hold true is when \(xy>{0}\) (so when \(x\) and \(y\) have the same sign) and \(|x|>|y|\) (simultaneously). In this case \(|x-y|={|x|-|y|}\) (as shown in my previous post). Example: \(x=3\) and \(y=2\) --> \(|x-y|=1={|x|-|y|}\); \(x=-3\) and \(y=-2\) --> \(|x-y|=1={|x|-|y|}\).

This post has left me confused. Earlier for these questions I used to plug in values and got to know if the inequality is valid or not.

As per the algebraic method, this inequality stands when

1. If x and y have different signs (got this by the plug in method) 2. If x and y have same signs but y>x (above post)

So now, we need to look out for a scenario when x and y are positive and y>x. So had the second statement been xy>0 then the answer would have been C?

Yes. If the 2nd statement had mentiond that xy>0 ---> this would mean that both x and y are of the same sign but without knowing whether y>x , this statement would not have been sufficient.

With statement 1, y<x and statement 2 (new) xy>0, you would have been able to answer the question unambiguously, leading to C.

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

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15 Oct 2015, 17:58

tricky one, as one combination of X and Y according to first statement gives makes the equation equal to whereas in question it is asked greater than, that the point which differentiates answer B from D
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

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

(1) y < x (2) xy < 0

If we modify the original condition, if |x|>=|y|, |x-y|>|x|-|y|>=0, so we can square both sides, which gives us (|x-y|)^2>(|x|-|y|)^2, and (x-y)^2>(|x|-|y|)^2 This becomes x^2+y^2-2xy>x^2+y^2-2|xy|, and if we simplify the inequality, -2xy>-2|xy|, and we ultimately want to know whether xy<0. This makes condition 2 sufficient. if |x|<|y|, then |x|-|y|<0, so |x-y|>|x|-|y| always work. Hence we do not need to deal with this, so the answer becomes (B).

Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.
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Re: Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 [#permalink]

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28 Oct 2015, 10:14

Alternate method:

|X-Y| > |X| - |Y|?

|x-y| => distance of y from X |x| => |x-0| => distance of x from zero |y| => |y-0| => distance of y from zero

Statement 1)

y< x

------0------y----x or ---y---0-------x---- or ----y------x-----0

distance still remains unknown ( not sufficient)

Statement 2)

----y---0----x--- or --x-----0----y

Distance between x and y will always be greater than (distance between x and 0) - (distance between y and 0) = > because of additive effect in the former and diminishing effect of the latter

Choice: B
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Re: Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 [#permalink]

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23 Jan 2016, 09:12

mmphf wrote:

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

(1) y < x (2) xy < 0

Let's just plug some numbers, I hate memorizing some complex formulas, in cases in which it's not needed. (1) x=10, y=3 --> 7 > 7 NO, now, let's make y negative: x=10, y=-3 --> 13 > 7 YES, two answers hence not sufficient

(2) two cases are possible: a) x- y+ b) x+ y-, --> there is no need to test case B, as we have already tested it above (x=10, y=-3 --> 13 > 7 YES) So, let's test case A: x=-5, y=5 --> 10 > 0 YES, we are done. Sufficient.

Answer B
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