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# Is |x-y|>|x|-|y|

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Is |x - y| > |x| - |y|? [#permalink]

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17 Nov 2009, 04:52
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Is |x - y| > |x| - |y|?

(1) y < x
(2) xy < 0
[Reveal] Spoiler: OA

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14 Jan 2012, 12:56
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mmphf wrote:
Is |x-y|>|x|-|y| ?

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

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

Probably the best way to solve this problem is plug-in method. Though there are two properties worth to remember:
1. Always true: $$|x+y|\leq{|x|+|y|}$$, note that "=" sign holds for $$xy\geq{0}$$ (or simply when $$x$$ and $$y$$ have the same sign);

2. Always true: $$|x-y|\geq{|x|-|y|}$$, note that "=" sign holds for $$xy>{0}$$ (so when $$x$$ and $$y$$ have the same sign) and $$|x|>|y|$$ (simultaneously). (Our case)

So, the question basically asks whether we can exclude "=" scenario from the second property.

(1) y < x --> we can not determine the signs of $$x$$ and $$y$$. Not sufficient.
(2) xy < 0 --> "=" scenario is excluded from the second property, thus $$|x-y|>|x|-|y|$$. Sufficient.

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13 Nov 2011, 17:29
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Is |x-y|>|x|-|y| ?

(1) y < x
(2) xy < 0
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17 Nov 2009, 06:28
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Basically the question asks whether the distance between the two points x and y on the line is greater than the difference between the individual distances of x and y from 0.

$$|x-y|>|x|-|y|$$?

(1) $$y<x$$, 3 possible cases for $$|x-y|>|x|-|y|$$:

A. ---------------$$0$$---$$y$$---$$x$$---, $$0<y<x$$ --> in this case $$|x-y|>|x|-|y|$$ becomes: $$x-y>x-y$$ --> $$0>0$$. Which is wrong;
B. ---------$$y$$---$$0$$---------$$x$$---, $$y<0<x$$ --> in this case $$|x-y|>|x|-|y|$$ becomes: $$x-y>x+y$$ --> $$y<0$$. Which is right, as we consider the range $$y<0<x$$;
C. ---$$y$$---$$x$$---$$0$$--------------, $$y<x<0$$ --> in this case $$|x-y|>|x|-|y|$$ becomes: $$x-y>-x+y$$ --> $$x>y$$. Which is right, as we consider the range $$y<x<0$$.

(2) $$xy<0$$, means $$x$$ and $$y$$ have different signs, hence 2 cases for $$|x-y|>|x|-|y|$$:

A. ----$$y$$-----$$0$$-------$$x$$---, $$y<0<x$$ --> in this case $$|x-y|>|x|-|y|$$ becomes: $$x-y>x+y$$ --> $$y<0$$. Which is right, as we consider the range $$y<0<x$$;
B. ----$$x$$-----$$0$$-------$$y$$---, $$x<0<y$$ --> in this case $$|x-y|>|x|-|y|$$ becomes: $$-x+y>-x-y$$ --> $$y>0$$. Which is right, as we consider the range $$x<0<y$$.

In both cases inequality holds true. Sufficient.

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25 Nov 2011, 15:00
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Here's how I did it.

|x-y| has a range of possible values , min being |x|-|y| and max being |x|+|y|

Statement 1 : x>y . Scenarioes :- x=+ve , y=-ve , |x-y|= |x|+|y| ;
x=+ve , y=+ve , |x-y|=|x|-|y| ;
x=-ve , y=-ve , |x-y|= |-(x-y)|=|x|-|y|

So we cannot definitely say that |x-y| is greater than |x|-|y| because the min value for |x-y| is also |x|-|y|. So, statement 1 is not sufficient.

Statement 2 : xy<0 . Scenarios :- x=+ve , y=-ve , |x-y|= |x|+|y|;
x=-ve , y=+ve , |x-y| = |-(|x|+|y|)|=|x|+|y|.

Now we know |x| + |y| is definitely greater than |x|-|y|. So statement (2) is sufficient.
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30 Nov 2011, 03:57
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Funny, but I remember form university that |a-b|>||a|-|b||>|a|-|b|, therefore the above inequality is valid for all numbers a,b can somebody verify the inequality?

http://math.ucsd.edu/~wgarner/math4c/de ... nequal.htm

If you notice, you have missed the 'equal to' sign.
Generalizing,$$|a-b|\geq|a|-|b|$$

In some cases, the equality will hold.
e.g. a = 3, b = 2
You get 1 = 1

In others, the inequality will hold.
e.g. a = -3, b = 4
7 > -1

In this question, you have to figure out whether the inequality will hold.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Math Expert Joined: 02 Sep 2009 Posts: 39725 Re: Is |x-y|>|x|-|y| [#permalink] ### Show Tags 14 Jan 2012, 13:18 4 This post received KUDOS Expert's post 4 This post was BOOKMARKED 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 $$y=-2$$ --> $$|x-y|=1={|x|-|y|}$$. Hope it's clear. _________________ Math Expert Joined: 02 Sep 2009 Posts: 39725 Re: gmat prep absolute values [#permalink] ### Show Tags 27 Jun 2010, 05:50 3 This post received KUDOS Expert's post 2 This post was BOOKMARKED SujD wrote: 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. _________________ Intern Status: Stay Hungry, Stay Foolish. Joined: 05 Sep 2011 Posts: 41 Location: India Concentration: Marketing, Social Entrepreneurship Re: Is |x-y|>|x|-|y| [#permalink] ### Show Tags 14 Nov 2011, 13:02 2 This post received KUDOS Statement 1) x>y. therefore, x-y>0. Plug & Play Method. (x,y)- (-3,-6) .Satisfies. (x,y)- (3,-6). Satisfies. (x-y)- (3,6). Does not Satisfy. Equality exists. Statement 2) xy<0. This means, either, x>0 and y<0. OR. x<0 and y>0. Looking the values plugged in statement 1. It satisfies the condition of statement two. Hence, B. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7447 Location: Pune, India Is |x-y|>|x|-|y| [#permalink] ### Show Tags 16 Nov 2014, 22:22 2 This post received KUDOS Expert's post mmphf wrote: Is |x-y|>|x|-|y| ? (1) y < x (2) xy < 0 Responding to a pm: You can solve this question easily if you understand some basic properties of absolute values. They are discussed in detail here: http://www.veritasprep.com/blog/2014/02 ... -the-gmat/ One of the properties is (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' Answer (B) _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Is |x - y | > |x | - |y | ? [#permalink]

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13 Aug 2016, 11:59
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Is |x - y | > |x | - |y | ?
(1) y < x
(2) xy < 0
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12 May 2010, 01:35
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LM wrote:
Please tell the quick approach.... it took me loner than I should have taken....

Algebraic approach is given in my first post. Below is another approach:

$$|x-y|>|x|-|y|$$?

(1) $$y<x$$

Try two positive number $$x=3>y=1$$ --> is $$|3-1|>|3|-|1|$$? --> is $$2>2$$? Answer NO.

Try ANY other case but both positive: $$x=-5>y=-7$$ --> is $$|-5-(-7)|>|-5|-|-7|$$? --> is $$2>-2$$? Answer YES.

(2) $$xy<0$$, means $$x$$ and $$y$$ have different signs.

Now we can spot here that when $$x$$ and $$y$$ have different signs $$x-y$$ always "contribute" to each other so that its absolute value will increase: $$x=3$$, $$y=-1$$ --> $$|x-y|=|3+1|=|4|=4$$ or $$x=-3$$, $$y=1$$ --> $$|x-y|=|-3-1|=|4|=4$$.

But $$|x|-|y|$$ is difference (thus not "contributing") of two positive values (as neither equals to zero). $$x=3$$, $$y=-1$$ --> $$|x|-|y|=|3|-|-1|=2$$ or $$x=-3$$, $$y=1$$ --> $$|x|-|y|=|-3|-|1|=2$$.

So $$xy<0$$ means $$|x-y|>|x|-|y|$$ is always true.

Sufficient.

Hope it helps.
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Re: Is |x – y| > |x| – |y|? [#permalink]

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18 Mar 2012, 01:46
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subhashghosh wrote:
(1) Y = -1, x = 0

Then, | 0 – (-1)| = 1
|0| - |-1| = -1

Y = 0, x = 1

Both |x-y| = |x| - |y| = 1

(2) Xy < 0, so one of them is < 0

So if we take the case x = -1, y = 1
Then |x – y| = |-2| = 2 and |x| - |y| = 1 – 1 = 0

Again, if x = 5 , y = -1
Then |x – y| = |6| = 6 and |x| - |y| = 5 – 1 = 4

So both 1 and 2 are insuff.

Combine them -> It is obvious that y < 0 and x > 0, so by adding a negative sign the magnitude increases and on the right side the magnitude will be less as the difference is between two positive numbers (i.e. the modulus values).

e.g. x = 2, y = -5

|x – y| = |7| and |x| - |y| = 2 – 5 = -3

So |x – y| > |x| - |y|

But in both the examples, its being shown that case 2 is sufficient. Am i mistaken?
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31 Mar 2012, 15:00
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I found talking through this one to be helpful.

Namely:

|x-y| represents the distance between x and y on the number line.

|x|-|y|, on the other hand, first takes the absolute value of both numbers - and thereby moving them both to the positive side of the number line - and THEN calculates the difference between x and y

Visually, it makes sense that if x and y are of different signs (for example, x=-5, y=5), then the difference between the two numbers on a number line is greater if measured before moving them both to the positive side of the number line.

At this point I logically deduced that it is impossible for |x-y| to be less than |x|-|y|. I also deduced at this point that if x and y have the same sign, it does not matter when the absolute value is taken because the difference between them will be the same either way.

After this thought process, the problem becomes much easier.
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25 Mar 2013, 05:29
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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

It is B because if you use the data of statement 2, you can say, "Yes, |x-y| is greater than |x|-|y|"

(2) xy < 0
This means that one of x and y is positive and the other is negative. You cannot take x = 1/2 and y = 1/3.
It is not about fractions/integers. It is about positive/negative numbers (most mod questions are about positive/negative numbers)
When xy < 0, |x-y|>|x|-|y| always holds.
Only when x and y both are positive or both are negative and |x|>|y|, then |x-y|=|x|-|y|
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Verbal Forum Moderator Joined: 10 Oct 2012 Posts: 629 Re: Is |x-y|>|x|-|y| [#permalink] ### Show Tags 25 Mar 2013, 12:39 1 This post received KUDOS 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 From F.S 1, we have that x>y. Thus |x-y| = x-y. Thus, we have to answer whether x-y>|x|-|y|. or x-|x|>y-|y|. Now for x>0, and y>0, we have is 0>0 and hence a NO. Again, for x>0 and y<0, we have a YES. Insufficient. For F.S 2, we know that x and y are of opposite signs. Thus, x and y being on the opposite sides of the number line w.r.t the origin, the term |x-y| will always be more than the difference of the absolute distance of x and y from origin.Sufficient. B. _________________ Last edited by mau5 on 05 Apr 2013, 05:02, edited 1 time in total. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7447 Location: Pune, India Re: Is |x-y|>|x|-|y| [#permalink] ### Show Tags 06 Aug 2013, 22:25 1 This post received KUDOS Expert's post 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 Fractions and integers have no role to play here. Check Bunuel's post above. Whenever xy < 0, i.e. x is negative but y is positive OR x is positive but y is negative, |x-y| is greater than |x|-|y|. e.g. x = -1/2, y = 1/3 |x-y| = |-1/2-1/3| = 5/6 |x|-|y| = 1/2 - 1/3 = 1/6 So |x - y| > |x|-|y| Do you see the logic here? If one of x and y is positive and the other is negative, in |x - y|, absolute values of x and y get added and the sum is positive. While in |x|-|y|, the absolute values are subtracted. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: Is |x - y| > |x| - |y|? [#permalink]

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01 Jan 2014, 21:55
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(1) is insuff. because if x and y equal to 10 and 2 respectively, then both sides of the inequality are the same weighted. But, if x, y are equal to 10 and -10 respectively, then left side is greater than the right side(from our perspective) of the inequality.

(2) states that either x or y is negative. We do not know which one is exactly negative. However, it is not important to find out this because in both cases the left side of the inequality is greater than the right side(from our perspective). Why?
It is simple. On the left side the numbers are added while on the right side the same numbers are subtracted from each other. Therefore, this statement is sufficient.

So, the correct answer is B.

Last edited by aja1991 on 02 Jan 2014, 03:36, edited 2 times in total.
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Re: Is |x - y| > |x| - |y|? [#permalink]

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24 Jan 2014, 14:17
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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

B.
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08 Jul 2014, 04:53
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Is this a valid approach to solve this problem?

| X –Y | > |X| - |Y|
Squaring both sides
(X-Y)^2>(|X|-|Y|)^2

X^2-2XY+Y^2>X^2-2|XY|+Y^2

-XY>|XY|

XY<|XY| --> Can be true only for XY < 0.

1 : y > X - Insufficient
2 : XY < 0 -> Sufficient.

Hence, (B).
Re: Is |x-y|>|x|-|y|   [#permalink] 08 Jul 2014, 04:53

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