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Re: Is |x-y|=||x|-|y||? (1) x>y (2) x<y<0 [#permalink]
16 May 2012, 00:32

Expert's post

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.

Re: Is |x-y|=||x|-|y||? (1) x>y (2) x<y<0 [#permalink]
16 May 2012, 02:04

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.

Re: Is |x-y|=||x|-|y||? (1) x>y (2) x<y<0 [#permalink]
16 May 2012, 02:08

Expert's post

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.

Answer: B.

Different approaches are possible to solve absolute value questions.

Re: Is |x-y|=||x|-|y||? (1) x>y (2) x<y<0 [#permalink]
27 Nov 2012, 18:12

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.

Re: Is |x-y|=||x|-|y||? (1) x>y (2) x<y<0 [#permalink]
28 Nov 2012, 03:36

Expert's post

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.

Answer: B.

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. _________________

Re: Is |x-y|=||x|-|y||? (1) x>y (2) x<y<0 [#permalink]
29 Nov 2012, 12:34

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.

Re: Is |x-y|=||x|-|y||? (1) x>y (2) x<y<0 [#permalink]
30 Nov 2012, 03:10

Expert's post

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. _________________

Re: Is |x-y|=||x|-|y||? (1) x>y (2) x<y<0 [#permalink]
30 Nov 2012, 04:26

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.

Answer: B.

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.

Bunnuel can you comment please. _________________

If you found my post useful and/or interesting - you are welcome to give kudos!

Re: Is |x-y|=||x|-|y||? (1) x>y (2) x<y<0 [#permalink]
30 Nov 2012, 04:30

Expert's post

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.

Answer: B.

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.

Bunnuel can you comment please.

You are right. The question is flawed in that respect. _________________

Re: Is |x-y|=||x|-|y||? (1) x>y (2) x<y<0 [#permalink]
10 Feb 2013, 06:33

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.

Answer: B.

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?

Re: Is |x-y|=||x|-|y||? (1) x>y (2) x<y<0 [#permalink]
23 Feb 2013, 13:06

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.

Answer: B.

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?

(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

Answer: B.

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

(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

Answer: B.

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

Please advice Cheers! J

Yes, your solution is perfectly fine. _________________

Re: Is |x-y| = ||x|-|y|| [#permalink]
04 Jun 2014, 19:26

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.

gmatclubot

Re: Is |x-y| = ||x|-|y||
[#permalink]
04 Jun 2014, 19:26

It has been a fairly long time since I have posted here, but I definitely did not want to sign off without giving readers a quick update on my personal...