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

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Is |x-y|>|x|-|y| [#permalink] New post 13 Nov 2011, 16:29
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Is |x-y|>|x|-|y| ?

(1) y < x
(2) xy < 0
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Re: Is |x-y|>|x|-|y| [#permalink] New post 14 Jan 2012, 11:56
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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.
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Re: Is |x-y|>|x|-|y| [#permalink] New post 14 Nov 2011, 12:02
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Statement 1)
x>y.
therefore, x-y>0.
Plug & Play Method.
(x,y)- (-3,-6) .Satisfies.
(x,y)- (3,-6). Satisfies.
(x-y)- (3,6). Does not Satisfy. Equality exists.


Statement 2)
xy<0.
This means,
either, x>0 and y<0.
OR. x<0 and y>0.

Looking the values plugged in statement 1.
It satisfies the condition of statement two.
Hence,
B.
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Re: Is |x-y|>|x|-|y| [#permalink] New post 25 Nov 2011, 14:00
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Answer is (B).

Here's how I did it.

|x-y| has a range of possible values , min being |x|-|y| and max being |x|+|y|

Statement 1 : x>y . Scenarioes :- x=+ve , y=-ve , |x-y|= |x|+|y| ;
x=+ve , y=+ve , |x-y|=|x|-|y| ;
x=-ve , y=-ve , |x-y|= |-(x-y)|=|x|-|y|

So we cannot definitely say that |x-y| is greater than |x|-|y| because the min value for |x-y| is also |x|-|y|. So, statement 1 is not sufficient.

Statement 2 : xy<0 . Scenarios :- x=+ve , y=-ve , |x-y|= |x|+|y|;
x=-ve , y=+ve , |x-y| = |-(|x|+|y|)|=|x|+|y|.

Now we know |x| + |y| is definitely greater than |x|-|y|. So statement (2) is sufficient.
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Re: Is |x-y|>|x|-|y| [#permalink] New post 30 Nov 2011, 02:57
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askou wrote:
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


If you notice, you have missed the 'equal to' sign.
Generalizing,|a-b|\geq|a|-|b|

In some cases, the equality will hold.
e.g. a = 3, b = 2
You get 1 = 1

In others, the inequality will hold.
e.g. a = -3, b = 4
7 > -1

In this question, you have to figure out whether the inequality will hold.
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Re: Is |x-y|>|x|-|y| [#permalink] New post 14 Jan 2012, 12:18
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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.
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Re: Is |x – y| > |x| – |y|? [#permalink] New post 18 Mar 2012, 00:46
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subhashghosh wrote:
(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


But in both the examples, its being shown that case 2 is sufficient. Am i mistaken?
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Re: Is |x-y|>|x|-|y| [#permalink] New post 31 Mar 2012, 14:00
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I found talking through this one to be helpful.

Namely:

|x-y| represents the distance between x and y on the number line.

|x|-|y|, on the other hand, first takes the absolute value of both numbers - and thereby moving them both to the positive side of the number line - and THEN calculates the difference between x and y

Visually, it makes sense that if x and y are of different signs (for example, x=-5, y=5), then the difference between the two numbers on a number line is greater if measured before moving them both to the positive side of the number line.

At this point I logically deduced that it is impossible for |x-y| to be less than |x|-|y|. I also deduced at this point that if x and y have the same sign, it does not matter when the absolute value is taken because the difference between them will be the same either way.

After this thought process, the problem becomes much easier.
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Re: Is |x-y|>|x|-|y| [#permalink] New post 25 Mar 2013, 04:29
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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 is B because if you use the data of statement 2, you can say, "Yes, |x-y| is greater than |x|-|y|"

(2) xy < 0
This means that one of x and y is positive and the other is negative. You cannot take x = 1/2 and y = 1/3.
It is not about fractions/integers. It is about positive/negative numbers (most mod questions are about positive/negative numbers)
When xy < 0, |x-y|>|x|-|y| always holds.
Only when x and y both are positive or both are negative and |x|>|y|, then |x-y|=|x|-|y|
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Re: Is |x-y|>|x|-|y| [#permalink] New post 06 Aug 2013, 21:25
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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


Fractions and integers have no role to play here. Check Bunuel's post above.

Whenever xy < 0, i.e. x is negative but y is positive OR x is positive but y is negative, |x-y| is greater than |x|-|y|.

e.g. x = -1/2, y = 1/3
|x-y| = |-1/2-1/3| = 5/6
|x|-|y| = 1/2 - 1/3 = 1/6

So |x - y| > |x|-|y|

Do you see the logic here? If one of x and y is positive and the other is negative, in |x - y|, absolute values of x and y get added and the sum is positive. While in |x|-|y|, the absolute values are subtracted.
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Re: Is |x-y|>|x|-|y| [#permalink] 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.
Cheers!
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Re: Is |x-y|>|x|-|y| [#permalink] 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] 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] New post 14 Jan 2012, 12:02
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] New post 14 Jan 2012, 13:52
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.


thanks. I missed that :)
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Re: Is |x-y|>|x|-|y| [#permalink] New post 19 Jan 2012, 07:48
Awesom explanation Bynuel..Kudos
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Re: Is |x – y| > |x| – |y|? [#permalink] 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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Re: Is |x – y| > |x| – |y|? [#permalink] New post 18 Mar 2012, 00:54
Sorry, was in a scatterbrained state then :), I think you're right. I just didn't do this problem in a focused manner.
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Is |x-y|>|x|-|y| ? [#permalink] 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] 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)
Re: Is |x - y | > |x | - |y | ? (1) y < x (2) xy < 0   [#permalink] 06 Mar 2013, 21:36
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