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Re: Is |x-y|>|x|-|y| [#permalink]

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New post 15 Oct 2015, 03:43
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Expert's post
longfellow wrote:
Bunuel wrote:
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.




Bunuel

This post has left me confused. Earlier for these questions I used to plug in values and got to know if the inequality is valid or not.

As per the algebraic method, this inequality stands when

1. If x and y have different signs (got this by the plug in method)
2. If x and y have same signs but y>x (above post)

So now, we need to look out for a scenario when x and y are positive and y>x. So had the second statement been xy>0 then the answer would have been C?


Yes. If the 2nd statement had mentiond that xy>0 ---> this would mean that both x and y are of the same sign but without knowing whether y>x , this statement would not have been sufficient.

With statement 1, y<x and statement 2 (new) xy>0, you would have been able to answer the question unambiguously, leading to C.
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Re: Is |x - y | > |x | - |y | ? [#permalink]

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New post 13 Aug 2016, 11:38
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Is |x - y | > |x | - |y | ?
(1) y < x
(2) xy < 0

Consider 1) y < x, let \(x = 1 , y = 0\)
then | x - y | = |1 - 0| = 1
| x | - | y | = |1| - |0| = 1 and 1 > 1 ; Therefore False
let \(x = 1 , y = -1\)
then | x - y | = 2
| x | - | y | = 0 and 2 > 0 ; Therefore True
Eliminate A and D, since it's inconsistent

Consider 2) xy < 0 , let \(x = 1 , y = -1\)
then | x - y | = |1 - -1| = 2
| x | - | y | = |1| - |1| = 0 and 2 > 1 ; Therefore True
let \(x = -1 , y = 1\)
then | x - y | = 2
| x | - | y | = 0 and 2 > 0 ; Therefore True
Consistent hence B
PS: The inequality need not be True or False, it just need to be consistent.
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Re: inequalities [#permalink]

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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]

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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]

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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: gmat prep absolute values [#permalink]

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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: Is |x-y|>|x|-|y| [#permalink]

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New post 13 Nov 2011, 17:45
mmphf wrote:
Is |x-y|>|x|-|y| ?

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



My Answer is E. Lets check this with pluy and play method.
Consider -
x= 5 and y = 2 -> 3 > 3
x= 2 and y = -2 -> 4 > 0
A not sufficient.

Consider -
x= 5 and y = 2 -> 3 > 3
x= -2 and y = -5 -> 3 > -3
B not sufficient.

Answer is E.
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Re: Is |x-y|>|x|-|y| [#permalink]

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New post 25 Nov 2011, 12:12
Capricorn369 wrote:
mmphf wrote:
Is |x-y|>|x|-|y| ?

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



My Answer is E. Lets check this with pluy and play method.
Consider -
x= 5 and y = 2 -> 3 > 3
x= 2 and y = -2 -> 4 > 0
A not sufficient.

Consider -
x= 5 and y = 2 -> 3 > 3
x= -2 and y = -5 -> 3 > -3
B not sufficient.

Answer is E.
Cheers!


Capricorn, in your explanation you plug in x=-2 and y=-5 for B. Check the problem one more time. Part B says that either x or y is negative, not both.
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Re: Is |x-y|>|x|-|y| [#permalink]

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New post 30 Nov 2011, 00:16
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
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Re: Is |x-y|>|x|-|y| [#permalink]

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New post 14 Jan 2012, 12:02
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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|
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Re: Is |x – y| > |x| – |y|? [#permalink]

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New post 17 Mar 2012, 23:56
(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|
Answer - C
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Is |x-y|>|x|-|y| ? [#permalink]

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New post 07 May 2012, 08:08
OA is B. I faced this problem in the GMATPrep.

(1) y < x
if x=3 and y=2, left hand abs(x-y) = 1, and right hand abs(x) - abs(y) = 1...No.
But if x=3 and y=-2, left hand is 5 and right is 1...Yes.
INSUFFICIENT.

(2) xy < 0
Let's think the following two cases.

(a) x>0 and y<0
In this case abs(x-y) > abs(x), as in the second plug-in in the discussion of (1) above.
So abs(x-y) naturally is greater than abs(x) - abs(y) because abs(x) > abs(x)-abs(y)...Yes.

(b) x<0 and y>0
In this case abs(x-y) = abs(x)+abs(y). Since abs(x) + abs(y) > abs (x) - abs(y),
abs(x-y) > abs(x)-abs(y)...Yes.

SUFFICIENT. :)

The answer is B.
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Re: Is |x - y | > |x | - |y | ? (1) y < x (2) xy < 0 [#permalink]

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New post 06 Mar 2013, 21:36
Answer: B

1) Taking statement (1), if y < x, then there can be two cases -
a) y is negative, it can lead to two sub cases --
(i) x is negative ==> as y < x so |y| > |x| ==> |x| - |y| will be < 0, and |x - y| > 0 ==> |x - y| > |x| - |y|
(ii) x is positive ==> |x - y| would be sum of absolute value of x and y, essentially |x| + |y| ==> |x - y| > |x| - |y|

Problem statement is true.

b) y is positive ==> x can only be positive ==> |x - y| = |x| - |y|

Problem statement is false.

Since we do not know, whether y is positive or negative we can not conclude from statement 1.

2) Taking statement (2), if xy < 0 ==> two sub cases
a) x < 0 and y > 0 ==> |x - y| = |x| + |y| which is greater than |x| - |y|
b) x > 0 and y < 0 ==> |x - y| = |x| + |y|, which is again greater than |x| - |y|

Statement (2) is sufficient enough to conclude the problem statement.

So Answer is (B)
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Re: Is |x-y|>|x|-|y| [#permalink]

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New post 08 Mar 2013, 04:56
So basically the explanation is that find the signs test in the following cases

++
- -
+ -
- +

since statement doesn't 1 doesn't help in any way its insufficient and statement 2 either both positive or both negative when we plug examples its never true so its sufficient?
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Re: Is |x-y|>|x|-|y| [#permalink]

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New post 25 Mar 2013, 02:08
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
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Re: Is |x-y|>|x|-|y| [#permalink]

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New post 25 Mar 2013, 03:56
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


The point is that x = 1/2 and y = 1/3 do not satisfy xy < 0 (the second statement).
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Re: Is |x-y|>|x|-|y| [#permalink]

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New post 25 Mar 2013, 10:19
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's implied that it is integers on the GMAT? Is this perception by me correct or completely out of the blue?
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Re: Is |x-y|>|x|-|y| [#permalink]

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New post 26 Mar 2013, 01:06
tulsa wrote:
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's implied that it is integers on the GMAT? Is this perception by me correct or completely out of the blue?


No, that's completely wrong, we cannot assume that x and y are integers, if this is not explicitly stated.

Generally, GMAT deals with only Real Numbers: Integers, Fractions and Irrational Numbers. So, if no limitations, then all we can say about a variable in a question that it's a real number.

For more check here: math-number-theory-88376.html

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

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New post 04 Apr 2013, 11:18
Bunuel wrote:
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.

Answer: B.


(1) x>y
x=-2,y=-4 then 2>-2 --> yes
x=4,y=-2 then 6>2 --> yes
can't get a no, so sufficient

(2) xy<0
x=4,y=-2 then 6>2 --> yes
x=-2,y=4 then 6>-2 --> yes
can't get a no, so sufficient

ans: D
why is the answer B? is the question mis-written and the inequality sign should have >= or <=?
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Re: Is |x-y|>|x|-|y| [#permalink]

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New post 05 Apr 2013, 01:55
margaretgmat wrote:
Bunuel wrote:
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.

Answer: B.


(1) x>y
x=-2,y=-4 then 2>-2 --> yes
x=4,y=-2 then 6>2 --> yes
can't get a no, so sufficient

(2) xy<0
x=4,y=-2 then 6>2 --> yes
x=-2,y=4 then 6>-2 --> yes
can't get a no, so sufficient

ans: D
why is the answer B? is the question mis-written and the inequality sign should have >= or <=?


What about the case x = 4, y = 2 in statement 1?
then we get 2 > 2 --> No
Hence statement 2 is not sufficient.
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Re: Is |x-y|>|x|-|y|   [#permalink] 05 Apr 2013, 01:55

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