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Re: Is |x-y|>|x|-|y| [#permalink]
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]
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?

Re: Is |x-y|>|x|-|y| [#permalink]
30 Nov 2011, 02:57

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Expert's post

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?

Re: Is |x-y|>|x|-|y| [#permalink]
14 Jan 2012, 11:56

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Expert's post

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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.

Re: Is |x-y|>|x|-|y| [#permalink]
14 Jan 2012, 12:18

3

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Expert's post

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|}.

Re: Is |x-y|>|x|-|y| [#permalink]
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|}.

Re: Is |x – y| > |x| – |y|? [#permalink]
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]
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?

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

Is |x-y|>|x|-|y| ? [#permalink]
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.

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

Re: Is |x-y|>|x|-|y| [#permalink]
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?

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