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

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Is |x - y| > |x| - |y|? [#permalink] New post 17 Nov 2009, 03:52
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Is |x - y| > |x| - |y|?

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

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Re: gmat prep absolute values [#permalink] New post 17 Nov 2009, 05:28
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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<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.
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Re: inequalities [#permalink] New post 14 Apr 2010, 06:38
raghavs wrote:
is |x-y|>|x|-|y|

1>y<x
2>x*y<0


IMO answer is 'B'

given expression, LHS = |x-y|; RHS = |x| - |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

OA pls
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Re: DS-Mode Inequality [#permalink] New post 11 May 2010, 13:32
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.

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.
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Re: Math (DS) [#permalink] New post 11 May 2010, 19:01
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.
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Re: Math (DS) [#permalink] New post 11 May 2010, 22:50
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?
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Re: DS-Mode Inequality [#permalink] New post 12 May 2010, 00:35
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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.

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.

Sufficient.

Answer: B.

Hope it helps.
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Re: Math (DS) [#permalink] New post 12 May 2010, 00:54
Hay Bunuel,
Thanks a lot... Now more clear and I accept your answer.
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Re: gmat prep absolute values [#permalink] New post 26 Jun 2010, 18:48
Bunuel wrote:
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?

Thanks for your help.
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Re: gmat prep absolute values [#permalink] New post 27 Jun 2010, 04:50
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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.
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Re: Is |x - y| > |x| - |y|? [#permalink] New post 19 Dec 2013, 06:01
study wrote:
Is |x - y| > |x| - |y|?

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


I think the trick here is to remember the property
|x - y| >= |x| - |y|

Where = will only happen when both x and y have the same sign
Therefore, we need to discard that case

Statement 2 does it

Hope it helps
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Re: Is |x - y| > |x| - |y|? [#permalink] New post 01 Jan 2014, 20:55
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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, 02:36, edited 2 times in total.
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Re: gmat prep absolute values [#permalink] New post 02 Jan 2014, 20:39
Bunuel wrote:
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.



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. :roll: :?:

please explain
thanks!
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Re: Is |x - y| > |x| - |y|? [#permalink] New post 02 Jan 2014, 22:36
study wrote:
Is |x - y| > |x| - |y|?

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


Statement I is insufficient

x....y.....(|x|-|y|)....|x-y|.....YES/NO
3....2.......1..............1........NO
3..-3........0..............6........YES

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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Re: gmat prep absolute values [#permalink] New post 03 Jan 2014, 03:40
Expert's post
rawjetraw wrote:
Bunuel wrote:
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.



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. :roll: :?:

please explain
thanks!


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

Hope it helps.
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PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Is |x - y| > |x| - |y|? [#permalink] New post 24 Jan 2014, 13:17
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.
Re: Is |x - y| > |x| - |y|?   [#permalink] 24 Jan 2014, 13:17
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