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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 \(y<0<x\).

Two different answers. Not sufficient.

(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 \(x<0<y\).

1) if y<x Case I: x<0 => y<0 --> LHS = RHS Case II: x>0, y>0 but <x --> LHS = RHS Case III: x>0, y<0 --> LHS > RHS Hence 1) alone is not sufficient

2) if x*y <0 => either x or y <0 and other has to be >0 Case I: x<0, y>0 --> LHS > RHS Cae II: x>0, y<0 --> LHS > RHS No other case. Hence, 2) alone is sufficient

So the only way that the absolute value of x-y is greater than the absolute value of x minus the absolute value of y could be true is if one or both is a negative number. Which one? Y i think. For example if x was 6 and y was -3. The left side would be l6--3l and the right l6l-l3l. Soto be true. it seems y should be negative. the firs statement says that y is less than x. Not sufficient to determine the initial question. If y is less than x but still positive the two sides are equal.

Second statement xy is less than o. This tells us that one number (x or y) is negative but not which one. Could this still be determinative? by itself no. but coupled with statement one we can say that x is positive and y is negative. In all such situations is the above question true. let's try x=4, y=-20. the right side l4--20l=24, the left side l4l-l20l= -16. Yes!

So I think both are needed. I have no idea how quick you were at solving this.

Okay so my analysis(as quoted below) on this question is wrong. Number two is sufficient. I see now that if either x or y is negative the right side will be greater.

is lx-yl > lxl - lyl ?

1. y<x

2. xy<o

So the only way that the absolute value of x-y is greater than the absolute value of x minus the absolute value of y could be true is if one or both is a negative number. Which one? Y i think. For example if x was 6 and y was -3. The left side would be l6--3l and the right l6l-l3l. Soto be true. it seems y should be negative. the firs statement says that y is less than x. Not sufficient to determine the initial question. If y is less than x but still positive the two sides are equal.

This is incorrectSecond statement xy is less than o. This tells us that one number (x or y) is negative but not which one. Could this still be determinative? by itself no. but coupled with statement one we can say that x is positive and y is negative. In all such situations is the above question true. let's try x=4, y=-20. the right side l4--20l=24, the left side l4l-l20l= -16. Yes!

So I think both are needed. I have no idea how quick you were at solving this.

BRAVO!!! skipjames BRAVO!!!! Some times when you look at any answer, your mind stops working its own way and start to accept what is stated. 1 kudo for you!! Can some body tell me how to give kudo?

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.

Two different answers. Not sufficient.

(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.

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 \(y<0<x\).

Two different answers. Not sufficient.

(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.

Answer: B.

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?

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 \(y<0<x\);

(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.

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.

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.

Hence the answer is B
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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 \(|x-y|=|-1-(-2)|=1=x-y\).

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

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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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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