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# Is |x-y| = ||x|-|y||

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05 Oct 2009, 09:52
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Is |x-y| = ||x|-|y||

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

Last edited by Bunuel on 03 Jul 2013, 06:03, edited 1 time in total.
Edited the question and added the OA.

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05 Oct 2009, 10:01
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Statement 1) Not sufficient.

Consider, x is -ve, y is -ve: x = -1, y = -2
|x-y| = 1, | |x|-|y| | = 1;

Consider, x is +ve, y is +ve: x = 2, y = 1
|x-y| = 1, | |x|-|y| | = 1;

Consider, x is +ve, y is -ve: x = 1, y = -1
|x-y| = 2, | |x|-|y| | = 0;

Statement 2) Sufficient.

Consider, x is -ve, y is -ve: x = -2, y = -1
|x-y| = 1, | |x|-|y| | = 1;

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05 Oct 2009, 10:44
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Expert's post
Economist wrote:
Is |x-y| = | |x|-|y| |
1). x>y
2). x<y<0

Is |x - y| = ||x| - |y|| ?

(1) x > y. There can be the following three cases:

A. x>y>0
LHS=x-y; RHS=|x-y|=x-y

B. x>0>y
LHS=x-y; RHS=|x+y|=-x-y or =x+y (depending whether |x|>|y| or not).

Already clear that (1) is not sufficient, but still let's continue:

C. 0>x>y
LHS=x-y; RHS=|-x+y|=x-y

NOT SUFFICIENT.

(2) x < y < 0:

LHS=-x+y; RHS=|-x+y|=-x+y.

SUFFICIENT.

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

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15 May 2012, 23:23
Is |x-y|=||x|-|y||?

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

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

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16 May 2012, 01:32
Is |x-y|=||x|-|y||?

(1) x>y. If x=1 and y=0 then the answer is YES but if x=1 and y=-1 then the answer is NO. Not sufficient.

(2) x<y<0. Analyze left hand side (LHS) of the equation: as x<y then x-y<0 so |x-y|=-(x-y)=y-x. Analyze right hand side (RHS) of the equation: as both x and y are negative then ||x|-|y||=|-x-(-y)|=|-x+y|=|y-x|. Again as x<y then y-x>0 so |y-x|=y-x. So, we have that LHS=RHS. Sufficient.

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

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16 May 2012, 03:04
Great. Thanks Bunuel for both the replies.

Is there any other approach to solve such questions?

Bunuel wrote:
Is |x-y|=||x|-|y||?

(1) x>y. If x=1 and y=0 then the answer is YES but if x=1 and y=-1 then the answer is NO. Not sufficient.

(2) x<y<0. Analyze left hand side (LHS) of the equation: as x<y then x-y<0 so |x-y|=-(x-y)=y-x. Analyze right hand side (RHS) of the equation: as both x and y are negative then ||x|-|y||=|-x-(-y)|=|-x+y|=|y-x|. Again as x<y then y-x>0 so |y-x|=y-x. So, we have that LHS=RHS. Sufficient.

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

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16 May 2012, 03:08
manjeet1972 wrote:
Great. Thanks Bunuel for both the replies.

Is there any other approach to solve such questions?

Bunuel wrote:
Is |x-y|=||x|-|y||?

(1) x>y. If x=1 and y=0 then the answer is YES but if x=1 and y=-1 then the answer is NO. Not sufficient.

(2) x<y<0. Analyze left hand side (LHS) of the equation: as x<y then x-y<0 so |x-y|=-(x-y)=y-x. Analyze right hand side (RHS) of the equation: as both x and y are negative then ||x|-|y||=|-x-(-y)|=|-x+y|=|y-x|. Again as x<y then y-x>0 so |y-x|=y-x. So, we have that LHS=RHS. Sufficient.

Different approaches are possible to solve absolute value questions.

DS questions on absolute value: search.php?search_id=tag&tag_id=37
PS questions on absolute value: search.php?search_id=tag&tag_id=58

Hope it helps.
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Re: Is |x-y|=||x|-|y||? (1) x>y (2) x<y<0 [#permalink]

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27 Nov 2012, 19:12
Bunuel,

I have noticed that you reason a lot of problems into their solution by simply picking a set of numbers. Other than positive, 0, negative and fractional numbers, do u follow some rule of thumb to directly see what numbers to plug. is there a more "right" number to plug in such that you arrive at solutions faster?
Any insight will be appreciated.

Bunuel wrote:
Is |x-y|=||x|-|y||?

(1) x>y. If x=1 and y=0 then the answer is YES but if x=1 and y=-1 then the answer is NO. Not sufficient.

(2) x<y<0. Analyze left hand side (LHS) of the equation: as x<y then x-y<0 so |x-y|=-(x-y)=y-x. Analyze right hand side (RHS) of the equation: as both x and y are negative then ||x|-|y||=|-x-(-y)|=|-x+y|=|y-x|. Again as x<y then y-x>0 so |y-x|=y-x. So, we have that LHS=RHS. Sufficient.

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

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28 Nov 2012, 04:36
koisun wrote:
Bunuel,

I have noticed that you reason a lot of problems into their solution by simply picking a set of numbers. Other than positive, 0, negative and fractional numbers, do u follow some rule of thumb to directly see what numbers to plug. is there a more "right" number to plug in such that you arrive at solutions faster?
Any insight will be appreciated.

Bunuel wrote:
Is |x-y|=||x|-|y||?

(1) x>y. If x=1 and y=0 then the answer is YES but if x=1 and y=-1 then the answer is NO. Not sufficient.

(2) x<y<0. Analyze left hand side (LHS) of the equation: as x<y then x-y<0 so |x-y|=-(x-y)=y-x. Analyze right hand side (RHS) of the equation: as both x and y are negative then ||x|-|y||=|-x-(-y)|=|-x+y|=|y-x|. Again as x<y then y-x>0 so |y-x|=y-x. So, we have that LHS=RHS. Sufficient.

First of all: on DS questions when plugging numbers, goal is to prove that the statement is not sufficient. So we should try to get a YES answer with one chosen number(s) and a NO with another.

Now, number picking strategy can vary for different problems. Generally it's good to test negative/positive/zero as well as integer/fraction to get a YES and a NO answers. If you deal with two variables it's also helpful to test x<y and x>y in addition to the former.
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Re: Is |x-y|=||x|-|y||? (1) x>y (2) x<y<0 [#permalink]

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29 Nov 2012, 13:34
I tried to solve it in this way, however I am not sure if my tght process is right. Bunuel, your input will be highly appreciated.

|x-y| = ||x|-|y||===> squaring both (given that both sides are mod)===> x2 + y2 -2xy = |x2|+|y2| -2|x||y|===> xy = |x||y|===> this is possible only when either both x,y >0 or x,y <0 .. second condition satisfies..hence B.

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

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30 Nov 2012, 04:10
pavanpuneet wrote:
I tried to solve it in this way, however I am not sure if my tght process is right. Bunuel, your input will be highly appreciated.

|x-y| = ||x|-|y||===> squaring both (given that both sides are mod)===> x2 + y2 -2xy = |x2|+|y2| -2|x||y|===> xy = |x||y|===> this is possible only when either both x,y >0 or x,y <0 .. second condition satisfies..hence B.

$$xy=|xy|$$ when $$xy\geq{0}$$. Apart from this your solution is correct.
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Re: Is |x-y|=||x|-|y||? (1) x>y (2) x<y<0 [#permalink]

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30 Nov 2012, 05:26
Bunuel wrote:
Is |x-y|=||x|-|y||?

(1) x>y. If x=1 and y=0 then the answer is YES but if x=1 and y=-1 then the answer is NO. Not sufficient.

(2) x<y<0. Analyze left hand side (LHS) of the equation: as x<y then x-y<0 so |x-y|=-(x-y)=y-x. Analyze right hand side (RHS) of the equation: as both x and y are negative then ||x|-|y||=|-x-(-y)|=|-x+y|=|y-x|. Again as x<y then y-x>0 so |y-x|=y-x. So, we have that LHS=RHS. Sufficient.

I am a little confused, as far as i understood in GMAT statements will not contradict to each other. But in this question in statement 1 x>y and vice versa in statement 2.

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

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30 Nov 2012, 05:30
ziko wrote:
Bunuel wrote:
Is |x-y|=||x|-|y||?

(1) x>y. If x=1 and y=0 then the answer is YES but if x=1 and y=-1 then the answer is NO. Not sufficient.

(2) x<y<0. Analyze left hand side (LHS) of the equation: as x<y then x-y<0 so |x-y|=-(x-y)=y-x. Analyze right hand side (RHS) of the equation: as both x and y are negative then ||x|-|y||=|-x-(-y)|=|-x+y|=|y-x|. Again as x<y then y-x>0 so |y-x|=y-x. So, we have that LHS=RHS. Sufficient.

I am a little confused, as far as i understood in GMAT statements will not contradict to each other. But in this question in statement 1 x>y and vice versa in statement 2.

You are right. The question is flawed in that respect.
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Re: Is |x-y|=||x|-|y||? (1) x>y (2) x<y<0 [#permalink]

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01 Dec 2012, 06:41
simple approach could be,

both sides should give an equal positive result

Consider x<y<0

x= -2 , y =-1

then,
|-2+1| = ||-2| - |-1||
|-1| = |2 - 1| since, modulus gives a positive result
1 = 1

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

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10 Feb 2013, 07:33
Bunuel wrote:
Is |x-y|=||x|-|y||?

(1) x>y. If x=1 and y=0 then the answer is YES but if x=1 and y=-1 then the answer is NO. Not sufficient.

(2) x<y<0. Analyze left hand side (LHS) of the equation: as x<y then x-y<0 so |x-y|=-(x-y)=y-x. Analyze right hand side (RHS) of the equation: as both x and y are negative then ||x|-|y||=|-x-(-y)|=|-x+y|=|y-x|. Again as x<y then y-x>0 so |y-x|=y-x. So, we have that LHS=RHS. Sufficient.

I tired to solve this question by using distances of x and y on the number line. My line of thinking was that I need to know where x and y sit with respect to zero. Is this a correct approach?

Thanks!

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

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23 Feb 2013, 14:06
I believe this approach is not correct. Your theory is correct that the distance between 2 numbers in the number line is the absolute value of the difference between the two numbers. However, the distance cannot always be measured as the difference of the absolute values as done in this problem. It could sometimes be the sum of the values also.

For example, consider distance between -5 and 3. The distance is |-5-5|=8
however, |-5|-|3|=2 which is wrong

alexpavlos wrote:
Bunuel wrote:
Is |x-y|=||x|-|y||?

(1) x>y. If x=1 and y=0 then the answer is YES but if x=1 and y=-1 then the answer is NO. Not sufficient.

(2) x<y<0. Analyze left hand side (LHS) of the equation: as x<y then x-y<0 so |x-y|=-(x-y)=y-x. Analyze right hand side (RHS) of the equation: as both x and y are negative then ||x|-|y||=|-x-(-y)|=|-x+y|=|y-x|. Again as x<y then y-x>0 so |y-x|=y-x. So, we have that LHS=RHS. Sufficient.

I tired to solve this question by using distances of x and y on the number line. My line of thinking was that I need to know where x and y sit with respect to zero. Is this a correct approach?

Thanks!

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

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30 Jun 2013, 11:07
Is |x-y|=||x|-|y||?

is x-y = |y|-|x|
OR
is x-y = |x|-|y|

(x-y)^2 = (|x|-|y|)^2?
(x-y)*(x-y) = (|x|-|y|)*(|x|-|y|)
x^2-2xy+y^2 = |x|^2 - 2|x|*|y| + |y|^2

-2xy = -2|xy|
xy = |xy|

That is only possible when xy = |xy| In other words, if xy were negative it wouldn't be equal to |xy|

(we cannot add 2xy to 2|xy| correct?)

(1) x>y

If x > y then |x-y| will always be positive. However, we don't know the sign of |x| and |y|
INSUFFICIENT

(2) x<y<0
x and y are both negative which means that xy = (-x)*(-y) which = (+xy)
SUFFICIENT

(B)

(just to make sure I understand it fully I will utilize Bunuel's approach as well)

#2) x<y<0
|x-y|=||x|-|y||
if x-y then |x-y| is negative: -(x-y) ===> (y-x)

if x and y are negative, then x, y are negative: (||x|-|y||) ===> | (-x) - (-y) | ===> |-x + y| ===> (y-x)
so: (y-x) = (y-x)
SUFFICIENT

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22 Apr 2014, 06:50
1
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Bunuel wrote:
Economist wrote:
Is |x-y| = | |x|-|y| |
1). x>y
2). x<y<0

(1)
A. x>y>0
LHS=x-y RHS=|x-y|=x-y
B. x>0>y
LHS=x-y RHS=|x+y|=-x-y or =x+y (depending |x|>|y| or not)
Already clear that (1) is not sufficient, but let's continue to make the way of dealing such problems more clear.
C. 0>x>y
LHS=x-y RHS=|-x+y|=x-y
NOT SUFFICINT

(2) x<y<0
LHS=-x+y RHS=|-x+y|=-x+y
SUFFICIENT

Is this approach valid?

Is |x-y| = | |x|-|y| |?

Square both sides and simplify
x^2-2xy+y^2 = x^2-2|x||y|+y^2?

We are down to is xy = |x||y|?

Statement 1 not sufficient
Statement 2 is sufficient and answer is thus yes

Therefore B stands

Cheers!
J

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22 Apr 2014, 07:35
jlgdr wrote:
Bunuel wrote:
Economist wrote:
Is |x-y| = | |x|-|y| |
1). x>y
2). x<y<0

(1)
A. x>y>0
LHS=x-y RHS=|x-y|=x-y
B. x>0>y
LHS=x-y RHS=|x+y|=-x-y or =x+y (depending |x|>|y| or not)
Already clear that (1) is not sufficient, but let's continue to make the way of dealing such problems more clear.
C. 0>x>y
LHS=x-y RHS=|-x+y|=x-y
NOT SUFFICINT

(2) x<y<0
LHS=-x+y RHS=|-x+y|=-x+y
SUFFICIENT

Is this approach valid?

Is |x-y| = | |x|-|y| |?

Square both sides and simplify
x^2-2xy+y^2 = x^2-2|x||y|+y^2?

We are down to is xy = |x||y|?

Statement 1 not sufficient
Statement 2 is sufficient and answer is thus yes

Therefore B stands

Cheers!
J

Yes, your solution is perfectly fine.
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Re: Is |x-y| = ||x|-|y|| [#permalink]

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04 Jun 2014, 20:26
Basically, the question is asking are both x and y +ve or both are -ve.
S1 does not give us any info regarding the signs i.e +ve or -ve.
S2 clearly states both are -ve.
Hence, B.

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Re: Is |x-y| = ||x|-|y||   [#permalink] 04 Jun 2014, 20:26
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